TÍCH PHÂN HÀM PHÂN THỨC HỮU TỈ CÓ MẪU LÀ ĐA THỨC BẬC CAO

Trong chuyên đề này, ta sẽ tìm hiểu các cách tính tích phân  $\int\limits_\alpha ^\beta  {\frac{{R\left( x \right)}}{{Q(x)}}dx}$ với Q(x) có bậc cao hơn 3.

Lưu ý: Đối với hàm phân thức hữu tỷ có bậc tử thấp hơn bậc mẫu tới hai bậc hoặc tinh ý nhận ra tính chất đặc biệt của hàm số dưới dấu tích phân thì ta có cách giải ngắn gọn hơn.

Ví dụ 1.
Tính các tích phân sau .
a. $\int\limits_1^2 {\frac{{dx}}{{x\left( {{x^4} + 1} \right)}}}$                b. $\int\limits_0^{\frac{1}{2}} {\frac{{{x^2} + 1}}{{{{\left( {x - 1} \right)}^3}\left( {x + 3} \right)}}dx\quad }$
Giải
a. $\int\limits_1^2 {\frac{{dx}}{{x\left( {{x^4} + 1} \right)}}}$ . Nếu theo cách phân tích bằng đồng nhất hệ số hai tử số thì ta có :
$f(x) = \frac{1}{{x\left( {{x^4} + 1} \right)}} = \frac{A}{x} + \frac{{B{x^3} + C{x^2} + Dx + E}}{{{x^4} + 1}} \\= \frac{{A\left( {{x^4} + 1} \right) + x\left( {B{x^3} + C{x^2} + Dx + E} \right)}}{{x\left( {{x^4} + 1} \right)}}$
$\Leftrightarrow f(x) = \frac{{\left( {A + B} \right){x^4} + C{x^3} + D{x^2} + {\text{Ex + A}}}}{{x\left( {{x^4} + 1} \right)}}\\ \Rightarrow \left\{ \begin{array}   A + B = 0  \\   C = 0,D = 0  \\   E = 0  \\   A = 1  \\ \end{array}  \right. \Leftrightarrow \left\{ \begin{array}   A = 1  \\   B =  - 1  \\   C = 0,D = 0,  \\   E = 0  \\ \end{array}  \right. \Rightarrow f(x) = \frac{1}{x} - \frac{{{x^3}}}{{{x^4} + 1}}$
Nhưng nếu ta tinh ý thì cách làm sau sẽ hay hơn .
Vì x và ${x^3}$ cách nhau 3 bậc , mặt khác $x \in \left[ {1;2} \right] \Rightarrow x \ne 0$. Cho nên ta nhân tử và mẫu với ${x^3} \ne 0$. Khi đó $f(x) = \frac{{{x^3}}}{{{x^4}\left( {{x^4} + 1} \right)}}$. Mặt khác $d\left( {{x^4}} \right) = 4{x^3}dx \Leftrightarrow dt = 4{x^3}dx\quad \left( {t = {x^4}} \right)$, cho nên :
$f(x)dx = \frac{1}{3}\frac{{3{x^3}dx}}{{{x^4}\left( {{x^4} + 1} \right)}} = \frac{1}{3}\frac{{dt}}{{t\left( {t + 1} \right)}} = \frac{1}{3}\left( {\frac{1}{t} - \frac{1}{{t + 1}}} \right) = f(t)$. Bài toán trở nên đơn giản hơn rất nhiều .
b. $\int\limits_0^{\frac{1}{2}} {\frac{{{x^2} + 1}}{{{{\left( {x - 1} \right)}^3}\left( {x + 3} \right)}}dx\quad }$
Nhận xét :
* Nếu theo phương pháp chung thì ta làm như sau :
- $f(x) = \frac{{{x^2} + 1}}{{{{\left( {x - 1} \right)}^3}\left( {x + 3} \right)}} = \frac{A}{{{{\left( {x - 1} \right)}^3}}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} + \frac{C}{{x - 1}} + \frac{D}{{x + 3}}$
- Sau đó quy đồng mẫu số , đồng nhất hệ số hai tử số , ta có : $A = \frac{1}{2},B = \frac{3}{8},C =  - D = \frac{5}{{32}}$
Do vậy : $I = \int\limits_0^{\frac{1}{2}} {\left( {\frac{1}{{2{{\left( {x - 1} \right)}^3}}} + \frac{3}{{8{{\left( {x - 1} \right)}^2}}} + \frac{5}{{32\left( {x - 1} \right)}} - \frac{5}{{32\left( {x + 3} \right)}}} \right)dx}$
$= \left[ { - \frac{1}{{8{{\left( {x - 1} \right)}^2}}} - \frac{3}{{8\left( {x - 1} \right)}} + \frac{5}{{32}}\ln \left| {x - 1} \right| - \frac{5}{{32}}\ln \left| {x + 3} \right|} \right]\left| {\begin{array}{*{20}{c}}   {\frac{1}{2}} \\   0 \end{array} = } \right.\frac{5}{{32}}\ln \frac{1}{{28}}$

Ví dụ 2.
Tính các tích phân sau :
a. $\int\limits_2^3 {\frac{{{x^4} - 1}}{{{x^6} - 1}}dx}$            b. $\int\limits_1^2 {\frac{{{x^2} + 1}}{{{x^6} + 1}}dx}$            c. $\int\limits_1^2 {\frac{{dx}}{{x\left( {1 + {x^4}} \right)}}}$
 d. $\int\limits_0^1 {\frac{{{x^3}}}{{{{\left( {1 + {x^2}} \right)}^3}}}dx}$        e. $\int\limits_0^1 {\frac{{{x^4} + 3{x^2} + 1}}{{{{\left( {1 + {x^2}} \right)}^3}}}dx}$        f. $\int\limits_{\frac{1}{3}}^1 {\frac{{{{\left( {x - {x^3}} \right)}^{\frac{1}{3}}}}}{{{x^4}}}dx}$
Giải
a. $\int\limits_1^2 {\frac{{{x^4} - 1}}{{{x^6} - 1}}dx}  = \int\limits_1^2 {\left( {\frac{{{x^4} + {x^2} + 1}}{{\left( {{x^2} - 1} \right)\left( {{x^4} + {x^2} + 1} \right)}} - \frac{{{x^2} + 2}}{{\left[ {{{\left( {{x^3}} \right)}^2} - 1} \right]}}} \right)dx}  = \int\limits_2^3 {\frac{1}{{{x^2} - 1}}dx + \int\limits_2^3 {\left( {\frac{{{x^2}}}{{\left[ {{{\left( {{x^3}} \right)}^2} - 1} \right]}} + \frac{1}{{{x^3} - 1}} - \frac{1}{{{x^3} + 1}}} \right)} } dx$
Tính J : J= artanx$\left| {\begin{array}{*{20}{c}}   3 \\   2 \end{array}} \right. = {\text{artan3 - artan2}}$.
Tính K . Đặt $t = {x^3} \Rightarrow \left\{ \begin{array}   dt = 3{x^2}dx,x = 2 \to t = 8;x = 3 \to t = 27  \\   g(x)dx = \frac{{{x^2}}}{{{x^3} - 1}}dx = \frac{1}{3}\frac{{dt}}{{\left( {{t^2} - 1} \right)}} = \frac{1}{3}\frac{1}{2}\left( {\frac{1}{{t - 1}} - \frac{1}{{t + 1}}} \right)dt  \\ \end{array}  \right.$
Do đó : K=$\int\limits_2^3 {g(x)dx}  = \frac{1}{6}\int\limits_8^{27} {\left( {\frac{1}{{t - 1}} - \frac{1}{{t + 1}}} \right)dt}  = \frac{1}{6}\left( {\ln \left| {t - 1} \right| - \ln \left| {t + 1} \right|} \right)\left| {\begin{array}{*{20}{c}}   {27} \\   8 \end{array} = \frac{1}{6}\ln \left| {\frac{{t - 1}}{{t + 1}}} \right|\left| {\begin{array}{*{20}{c}}   {27} \\   8 \end{array} = } \right.} \right.\frac{1}{6}\ln \frac{{117}}{{98}}$
Tính E=$\int\limits_2^3 {\frac{1}{{{x^3} - 1}}dx}  = \int\limits_2^3 {\frac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}dx}$
Ta có : $h(x) = \frac{1}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{{x^2} - \left( {{x^2} - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{{x^2}}}{{{x^3} - 1}} - \frac{{{x^2} - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$
$= \frac{{{x^2}}}{{{x^3} - 1}} - \frac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{{x^2}}}{{{x^3} - 1}} - \frac{{x + 1}}{{{x^2} + x + 1}} = \frac{{{x^2}}}{{{x^3} - 1}} - \frac{1}{2}\left( {\frac{{2x + 1}}{{{x^2} + x + 1}} + \frac{1}{{{x^2} + x + 1}}} \right)$
Vậy : $I = \frac{1}{3}\int\limits_2^3 {\frac{{3{x^2}}}{{{x^3} - 1}}dx}  - \frac{1}{2}\int\limits_2^3 {\frac{{\left( {2x + 1} \right)}}{{{x^2} + x + 1}}dx - \int\limits_2^3 {\frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}dx} }$
$= \frac{1}{3}\ln \left( {{x^3} - 1} \right)\left| {\begin{array}{*{20}{c}}   3 \\   2 \end{array} - \frac{1}{2}\ln \left( {{x^2} + x + 1} \right)\left| {\begin{array}{*{20}{c}}   3 \\   2 \end{array} - F = \frac{1}{3}\ln \frac{{28}}{9} - \frac{1}{2}\ln \frac{{13}}{6} - F} \right.} \right.\quad \left( 2 \right)$
Tính F: Đặt : $x + \frac{1}{2} = \frac{{\sqrt 3 }}{2}\tan t \Rightarrow \left\{ \begin{array}   dx = \frac{{\sqrt 3 }}{2}\frac{1}{{c{\text{o}}{{\text{s}}^{\text{2}}}t}}dt  \\   x = 2 \to \tan t = \frac{5}{{\sqrt 3 }} \to t = a;x = 3 \to \tan t = \frac{{10}}{{\sqrt 3 }} \to t = b  \\ \end{array}  \right.$
Do đó F=$\int\limits_a^b {\frac{{\frac{{\sqrt 3 }}{2}\frac{1}{{c{\text{o}}{{\text{s}}^{\text{2}}}t}}dt}}{{\frac{{\sqrt 3 }}{2}\left( {1 + {{\tan }^2}t} \right)}} = \int\limits_a^b {dt}  = t\left| {\begin{array}{*{20}{c}}   b \\   a \end{array} = b - a\quad \left( {\operatorname{t} {\text{ant = }}\frac{{\text{5}}}{{\sqrt {\text{3}} }} \to t = a = {\text{artan}}\frac{{\text{5}}}{{\sqrt {\text{3}} }};b = {\text{artan}}\frac{{{\text{10}}}}{{\sqrt {\text{3}} }}} \right)} \right.}$
Thay vào (2) ta có kết quả .
b. $\int\limits_1^2 {\frac{{{x^2} + 1}}{{{x^6} + 1}}dx}  = \int\limits_0^1 {\frac{{{x^2} + 1}}{{\left( {{x^2} + 1} \right)\left( {{x^4} - {x^2} + 1} \right)}}dx = \int\limits_1^2 {\frac{1}{{{{\left( {{x^2} - 1} \right)}^2} - {x^2}}}dx} }  = \int\limits_1^2 {\frac{1}{{\left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)}}dx}$
Ta có : $\frac{1}{{\left( {{x^2} + x + 1} \right)\left( {{x^2} - x + 1} \right)}} = \frac{{{\text{Ax + B}}}}{{{x^2} + x + 1}} + \frac{{Cx + D}}{{{x^2} - x + 1}}$
$= \frac{{\left( {A + C} \right){x^3} + \left( {B - A + C + D} \right){x^2} + \left( {A - B + C + D} \right)x + \left( {B + D} \right)}}{{{x^4} - {x^2} + 1}}$
Đồng nhất hệ số hai tử số ta có hệ : $\left\{ \begin{array}   A + C = 0  \\   B - A + C + D = 0  \\   A - B + C + D = 0  \\   B + D = 1  \\ \end{array}  \right. \Leftrightarrow \left\{ \begin{array}   A =  - C  \\   1 - 2C = 0  \\    - B + D = 0  \\   B + D = 1  \\ \end{array}  \right. \Leftrightarrow \left\{ \begin{array}   A =  - \frac{1}{2}  \\   C = \frac{1}{2}  \\   D = \frac{1}{2}  \\   B = \frac{1}{2}  \\ \end{array}  \right.$
Vậy : $I = \frac{1}{2}\left( {\int\limits_1^2 {\frac{{1 - x}}{{{x^2} + x + 1}}dx + \int\limits_1^2 {\frac{{x + 1}}{{{x^2} - x + 1}}dx} } } \right) = \frac{1}{2}\left( {J + K} \right)\left( 1 \right)$
Tính J=$\int\limits_1^2 {\frac{{ - x + 1}}{{{x^2} + x + 1}}dx}  =  - \frac{1}{2}\int\limits_1^2 {\frac{{2x + 1 - 3}}{{{x^2} + x + 1}}dx}$
$=  - \frac{1}{2}\int\limits_1^2 {\frac{{2x + 1}}{{{x^2} + x + 1}}dx}  + \frac{3}{2}\int\limits_1^2 {\frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}dx}  =  - \frac{1}{2}\ln \left| {{x^2} + x + 1} \right|\left| {\begin{array}{*{20}{c}}   2 \\   1 \end{array} + E\quad \left( 2 \right)} \right.$
Tính E =$\frac{3}{2}\int\limits_1^2 {\frac{1}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}dx}$, (đặt : $x + \frac{1}{2} = \frac{{\sqrt 3 }}{2}\tan t$)
Tính K
$K = \int\limits_1^2 {\frac{{x + 1}}{{{x^2} - x + 1}}dx}  = \frac{1}{2}\int\limits_1^2 {\frac{{2x - 1 + 3}}{{{x^2} - x + 1}}dx} \\ = \frac{1}{2}\int\limits_1^2 {\frac{{2x - 1}}{{{x^2} - x + 1}}dx}  + \frac{3}{2}\int\limits_0^1 {\frac{1}{{{{\left( {x - \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}dx = \frac{1}{2}\ln \left| {{x^2} - x + 1} \right|\left| {\begin{array}{*{20}{c}}   2 \\   1 \end{array} + F\quad \left( 2 \right)} \right.}$
Tính F=$\frac{3}{2}\int\limits_1^2 {\frac{1}{{{{\left( {x - \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}dx}$, (đặt $x - \frac{1}{2} = \frac{{\sqrt 3 }}{2}\tan t$)
c. $\int\limits_1^2 {\frac{{dx}}{{x\left( {1 + {x^4}} \right)}}}  = \frac{1}{3}\int\limits_1^2 {\frac{{3{x^3}}}{{{x^4}\left( {1 + {x^4}} \right)}}dx}  = \frac{1}{3}\int\limits_1^2 {\left( {\frac{{d\left( {{x^4}} \right)}}{{{x^4}}} - \frac{{d\left( {{x^4}} \right)}}{{1 + {x^4}}}} \right) = \frac{1}{3}\ln \left( {\frac{{{x^4}}}{{1 + {x^4}}}} \right)\left| {\begin{array}{*{20}{c}}   2 \\   1 \end{array} = \frac{1}{3}\ln \frac{{32}}{{17}}} \right.}$
d. $\int\limits_0^1 {\frac{{{x^3}}}{{{{\left( {1 + {x^2}} \right)}^3}}}dx}  = \frac{1}{2}\int\limits_0^1 {\frac{{{x^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}}2xdx} \quad \left( 1 \right)$. Đặt : $t = 1 + {x^2} \Rightarrow \left\{ \begin{array}   {x^2} = t - 1;dt = 2xdx  \\   x = 0 \to t = 1,x = 1 \to t = 2  \\ \end{array}  \right.$
Do đó $I = \int\limits_1^2 {\frac{{t - 1}}{{{t^3}}}dt}  = \int\limits_1^2 {\left( {\frac{1}{{{t^2}}} - \frac{1}{{{t^3}}}} \right)dt}  = \left( { - \frac{1}{t} + \frac{1}{{4{t^2}}}} \right)\left| {\begin{array}{*{20}{c}}   2 \\   1 \end{array} = \frac{{13}}{{16}}} \right.$
e. $\int\limits_0^1 {\frac{{{x^4} + 3{x^2} + 1}}{{{{\left( {1 + {x^2}} \right)}^3}}}dx}  = \int\limits_0^1 {\left( {\frac{{{{\left( {1 + {x^2}} \right)}^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}} + \frac{{{x^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}}} \right)dx}  = \int\limits_0^1 {\frac{1}{{1 + {x^2}}}dx}  + \int\limits_0^1 {\frac{{{x^2}}}{{{{\left( {1 + {x^2}} \right)}^3}}}dx}  = J + K\left( 1 \right)$
Tính J : Bằng cách đặt $x = \tan t \Rightarrow J = \frac{\pi }{4}$
Tính K=$\int\limits_0^1 {\left( {\frac{1}{{{{\left( {1 + {x^2}} \right)}^2}}} - \frac{1}{{{{\left( {1 + {x^2}} \right)}^3}}}} \right)dx}  = E + F\left( 2 \right)$
Tính E : Bằng cách đặt $\begin{array}   x = \tan t \leftrightarrow \left\{ \begin{array}   dx = \frac{1}{{c{\text{o}}{{\text{s}}^{\text{2}}}t}}dt  \\   x = 0 \to t = 0;x = 1 \to t = \frac{\pi }{4}  \\ \end{array}  \right.  \\     \\ \end{array}$
Vậy : $E = \frac{1}{2}\int\limits_0^1 {{{\left( {\frac{1}{{1 + {x^2}}}} \right)}^2}dx = \frac{1}{2}} \int\limits_0^{\frac{\pi }{4}} {{{\left( {\frac{1}{{1 + {{\tan }^2}t}}} \right)}^2}\frac{1}{{c{\text{o}}{{\text{s}}^{\text{2}}}t}}dt = } \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\frac{1}{{\frac{1}{{c{\text{o}}{{\text{s}}^{\text{4}}}t}}}}\frac{1}{{c{\text{o}}{{\text{s}}^{\text{2}}}t}}dt}  = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {c{\text{o}}{{\text{s}}^{\text{2}}}tdt}$
$= \frac{1}{4}\int\limits_0^{\frac{\pi }{4}} {\left( {1 + c{\text{os2t}}} \right)dt}  = \frac{1}{4}\left( {t + \frac{1}{2}\sin 2t} \right)\left| {\begin{array}{*{20}{c}}   {\frac{\pi }{4}} \\   0 \end{array} = \frac{1}{4}\left( {\frac{\pi }{4} + \frac{1}{2}} \right) = \frac{{\pi  + 2}}{{16}}} \right.$
Tính F. Tương tự như tính E ;
Bằng cách đặt $\begin{array}   x = \tan t \leftrightarrow \left\{ \begin{array}   dx = \frac{1}{{c{\text{o}}{{\text{s}}^{\text{2}}}t}}dt  \\   x = 0 \to t = 0;x = 1 \to t = \frac{\pi }{4}  \\ \end{array}  \right.  \\     \\ \end{array}$
Vậy : $F = \frac{1}{2}\int\limits_0^1 {{{\left( {\frac{1}{{1 + {x^2}}}} \right)}^3}dx = \frac{1}{2}} \int\limits_0^{\frac{\pi }{4}} {{{\left( {\frac{1}{{1 + {{\tan }^2}t}}} \right)}^3}\frac{1}{{c{\text{o}}{{\text{s}}^{\text{2}}}t}}dt = } \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {\frac{1}{{\frac{1}{{c{\text{o}}{{\text{s}}^{\text{6}}}t}}}}\frac{1}{{c{\text{o}}{{\text{s}}^{\text{2}}}t}}dt}  = \frac{1}{2}\int\limits_0^{\frac{\pi }{4}} {c{\text{o}}{{\text{s}}^{\text{4}}}tdt}$
$= \frac{1}{8}\int\limits_0^{\frac{\pi }{4}} {{{\left( {1 + c{\text{os2t}}} \right)}^2}dt}  = \frac{1}{8}\int\limits_0^{\frac{\pi }{4}} {\left( {1 + 2c{\text{os}}2t + \frac{{1 + c{\text{os4t}}}}{2}} \right)} dt\left| {\begin{array}{*{20}{c}}   {\frac{\pi }{4}} \\   0 \end{array} = } \right.$
$\frac{1}{{16}}\int\limits_0^{\frac{\pi }{4}} {\left( {3 + 4\cos 2t + c{\text{os4t}}} \right)dt = } \frac{1}{{16}}\left( {3t + 2\sin 2t + \frac{1}{4}\sin 4t} \right)\left| {\begin{array}{*{20}{c}}   {\frac{\pi }{4}} \\   0 \end{array} = \frac{1}{{16}}\left( {3\frac{\pi }{4} + 2} \right) = \frac{{3\pi  + 8}}{{64}}} \right.$
f. $\int\limits_{\frac{1}{3}}^1 {\frac{{{{\left( {x - {x^3}} \right)}^{\frac{1}{3}}}}}{{{x^4}}}dx}  = \int\limits_{\frac{1}{3}}^1 {{{\left( {\frac{{x - {x^3}}}{{{x^3}}}} \right)}^{\frac{1}{3}}}\frac{1}{{{x^3}}}dx}  = \int\limits_{\frac{1}{3}}^1 {{{\left( {\frac{1}{{{x^2}}} - 1} \right)}^{\frac{1}{3}}}\frac{1}{{{x^2}}}.\frac{{dx}}{x}}$
Đặt : $t = \left( {\frac{1}{{{x^2}}} - 1} \right) \Rightarrow t + 1 = \frac{1}{{{x^2}}} \Leftrightarrow \left\{ \begin{array}   dt =  - \frac{{dx}}{x}  \\   x = \frac{1}{3} \to t = 8;x = 1 \to t = 0  \\ \end{array}  \right.$
Khi đó $I =  - \int\limits_8^0 {{t^{\frac{1}{3}}}\left( {t + 1} \right)dt}  = \int\limits_0^8 {\left( {{t^{\frac{4}{3}}} + {t^{\frac{1}{3}}}} \right)dt}  = \left( {\frac{3}{7}{t^{\frac{7}{3}}} + \frac{3}{4}{t^{\frac{4}{3}}}} \right)\left| {\begin{array}{*{20}{c}}   8 \\   0 \end{array} = \frac{3}{7}{{.2}^7} + \frac{3}{4}{{.2}^4} = 16\left( {\frac{{24}}{7} + \frac{3}{4}} \right) = \frac{{468}}{7}} \right.$

Ví dụ 3.
Tính các tích phân sau
a.$\int\limits_1^{{e^{\frac{1}{{p + 2}}}}} {\frac{{{x^{\frac{p}{2}}}}}{{{x^{p + 2}} + 1}}dx}$                b. $\int\limits_0^a {\frac{{{x^3}dx}}{{{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}}}$
   c. $\int\limits_0^1 {{e^{x + {e^x}}}dx}$                    d. $\int\limits_0^{2a} {x\sqrt {2ax - {x^2}} dx}$
Giải
a. $\int\limits_1^{{e^{\frac{1}{{p + 2}}}}} {\frac{{{x^{\frac{p}{2}}}}}{{{x^{p + 2}} + 1}}dx}$ :  Ta có : $f(x)dx = \frac{{{x^{\frac{p}{2}}}dx}}{{{{\left( {{x^{\frac{{p + 2}}{2}}}} \right)}^2} + 1}}$.
- Đặt : $t = {x^{\frac{{p + 2}}{2}}} = {x^{\frac{p}{2} + 1}} \Rightarrow \left[ \begin{array}   dt = {x^{\frac{p}{2}}}dx  \\   x = 1 \to t = 1;x = {e^{\frac{1}{{p + 2}}}} \to t = \sqrt e   \\ \end{array}  \right. \Leftrightarrow I = \int\limits_1^{\sqrt e } {\frac{{dt}}{{{t^2} + 1}}}$   
- Đặt : $t = \tan u \Rightarrow \left[ \begin{array}   dt = \frac{{du}}{{c{\text{o}}{{\text{s}}^{\text{2}}}u}}  \\   t = 1 \to u = \frac{\pi }{4},t = {e^{\frac{1}{2}}} \to u = {u_1}  \\ \end{array}  \right. \Leftrightarrow I = \int\limits_{\frac{\pi }{4}}^{{u_1}} {\frac{{du}}{{c{\text{o}}{{\text{s}}^{\text{2}}}u\left( {1 + {{\tan }^2}u} \right)}} = \int\limits_{\frac{\pi }{4}}^{{u_1}} {du = \frac{\pi }{4} - {u_1}} }$
- Từ : $\tan u = \sqrt e  \Rightarrow u = {u_1} = {\text{artan}}\sqrt {\text{e}}  \Leftrightarrow I = \frac{\pi }{4} - {\text{artan}}\sqrt {\text{e}}$
b. $\int\limits_0^a {\frac{{{x^3}dx}}{{{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}}}$.
Đặt : $x = {\text{atant}} \Rightarrow \left\{ \begin{array}   {\text{dx = a}}\frac{{{\text{dt}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}t}};x = 0 \to t = 0,x = a \to t = \frac{\pi }{4}  \\   f(x) = \frac{{{x^3}dx}}{{{{\left( {{x^2} + {a^2}} \right)}^{\frac{3}{2}}}}} = \frac{{{a^3}{{\tan }^3}t}}{{{a^3}{{\left( {\frac{1}{{c{\text{o}}{{\text{s}}^{\text{2}}}t}}} \right)}^{\frac{3}{2}}}}}a\frac{{{\text{dt}}}}{{{\text{co}}{{\text{s}}^{\text{2}}}t}} = a\cos t.{\tan ^3}tdt  \\ \end{array}  \right.$
Vậy : $I = \int\limits_0^a {f(x)dx}  = \int\limits_0^{\frac{\pi }{4}} {a\cos t.{{\tan }^3}tdt}  = \int\limits_0^{\frac{\pi }{4}} {a\cos t.\frac{{{{\sin }^3}t}}{{c{\text{o}}{{\text{s}}^{\text{3}}}t}}dt}  = \int\limits_0^{\frac{\pi }{4}} {a.\frac{{{{\sin }^3}t}}{{c{\text{o}}{{\text{s}}^{\text{2}}}t}}dt = } a\int\limits_0^{\frac{\pi }{4}} {\frac{{\left( {1 - c{\text{o}}{{\text{s}}^{\text{2}}}t} \right)\sin t}}{{c{\text{o}}{{\text{s}}^{\text{2}}}t}}dt}$
- Đặt : $c{\text{ost = u}} \Rightarrow \left\{ \begin{array}   du =  - \operatorname{s} {\text{intdt;t = }}\frac{\pi }{4} \to u = \frac{1}{{\sqrt 2 }};t = 0 \to u = 1  \\   f(t)dt = \frac{{\left( {1 - {u^2}} \right)}}{{{u^2}}}\left( { - du} \right) = \left( {1 - \frac{1}{{{u^2}}}} \right)du  \\ \end{array}  \right.$
Vậy : $I = \int\limits_1^{\frac{{\sqrt 2 }}{2}} {\left( {1 - \frac{1}{{{u^2}}}} \right)du = \left( {u + \frac{1}{u}} \right)\left| {\begin{array}{*{20}{c}}   {\frac{{\sqrt 2 }}{2}} \\   1 \end{array} = \frac{{\sqrt 2 }}{2} + \frac{2}{{\sqrt 2 }} - 2 = \frac{3}{{\sqrt 2 }} - 2 = \frac{{3\sqrt 2 }}{2} - 2 = \frac{{3\sqrt 2  - 4}}{2}} \right.}$
c. $\int\limits_0^1 {{e^{x + {e^x}}}dx}  = \int\limits_0^1 {{e^x}{e^{{e^x}}}dx}$. Đặt : $t = {e^x} \Rightarrow \left\{ \begin{array}   dt = {e^x}dx;x = 0 \to t = 1;x = 1 \to t = e  \\   f(x)dx = {e^x}{e^{{e^x}}}dx = {e^t}dt  \\ \end{array}  \right.$
Vậy : $I = \int\limits_0^1 {f(x)dx}  = \int\limits_1^e {{e^t}dt}  = {e^t}\left| {\begin{array}{*{20}{c}}   e \\   1 \end{array} = {e^e} - e} \right.$
d. $\int\limits_0^{2a} {x\sqrt {2ax - {x^2}} dx}  = \int\limits_0^{2a} {x\sqrt {{a^2} - {{\left( {x - a} \right)}^2}} dx}$
Đặt : $x - a = a.\sin t \Rightarrow \left\{ \begin{array}   dx = a.c{\text{ostdt,x = 0}} \to {\text{t =  - }}\frac{\pi }{2}{\text{;x = 2a}} \to {\text{t = }}\frac{\pi }{2}  \\   f(x)dx = \left( {a + a.\sin t} \right)\sqrt {{a^2}c{\text{o}}{{\text{s}}^{\text{2}}}t} .a.c{\text{ostdt}}  \\ \end{array}  \right.$
Vậy : $I = {a^3}\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left( {1 + \sin t} \right)c{\text{o}}{{\text{s}}^{\text{2}}}tdt}  = {a^3}\left[ {\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {c{\text{o}}{{\text{s}}^{\text{2}}}tdt + \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {c{\text{o}}{{\text{s}}^{\text{2}}}t\sin tdt} } } \right] = {a^3}\left[ {\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{1 + c{\text{os2}}t}}{2}dt - \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {c{\text{o}}{{\text{s}}^{\text{2}}}td\left( {c{\text{os}}t} \right)} } } \right]$
$= {a^3}\left[ {\frac{1}{2}\left( {t + \frac{1}{2}\sin 2t} \right)\left| {\begin{array}{*{20}{c}}   {\frac{\pi }{2}} \\   { - \frac{\pi }{2}} \end{array} - \frac{1}{3}{\text{co}}{{\text{s}}^{\text{3}}}t\left| {\begin{array}{*{20}{c}}   {\frac{\pi }{2}} \\   { - \frac{\pi }{2}} \end{array}} \right.} \right.} \right] = {a^3}\left[ {\frac{1}{2}\left( {\frac{\pi }{2} + \frac{\pi }{2}} \right)} \right] = {a^3}\frac{\pi }{2}$

Ví dụ 4.

Tính các tích phân sau
a. $\int\limits_2^3 {\frac{{dx}}{{{x^5} - {x^2}}}}$                    b. $\int\limits_0^1 {\frac{{{x^7}dx}}{{{{\left( {1 + {x^4}} \right)}^2}}}}$
 c. $\int\limits_0^1 {\frac{{{x^3} - 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx}$                d. $\int\limits_1^2 {\frac{{\sqrt {1 + {x^3}} }}{{{x^4}}}dx}$
Giải
a. $\int\limits_2^3 {\frac{{dx}}{{{x^5} - {x^2}}}}  = \int\limits_2^3 {\frac{1}{{{x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}dx\quad \left( 1 \right)}$
Xét : $f(x) = \frac{1}{{{x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{A}{{{x^2}}} + \frac{B}{x} + \frac{{Cx + D}}{{{x^2} + x + 1}} + \frac{E}{{x - 1}}$
$= \frac{{A\left( {{x^2} + x + 1} \right)\left( {x - 1} \right) + Bx\left( {x - 1} \right)\left( {{x^2} + x + 1} \right) + \left( {Cx + D} \right){x^2}\left( {x - 1} \right) + E({x^2} + x + 1){x^2}}}{{{x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$
$= \frac{{\left( {B + C + E} \right){x^4} + \left( {A + D - C + E} \right){x^3} + \left( {E - D} \right){x^2} - Bx - A}}{{{x^2}\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$.
Đồng nhất hệ số hai tử số ta có hệ :
$\left\{ \begin{array}   B + C + E = 0  \\   A + D - C + E = 0  \\   E - D = 0  \\   B = 0  \\   A =  - 1  \\ \end{array}  \right. \Leftrightarrow \left\{ \begin{array}   C =  - E  \\   E + E + E = 1  \\   B = 0  \\   E = D  \\   A =  - 1  \\ \end{array}  \right. \Leftrightarrow \left\{ \begin{array}   D = \frac{1}{3}  \\   C =  - \frac{1}{3}  \\   B = 0  \\   E = \frac{1}{3}  \\   A =  - 1  \\ \end{array}  \right. \Rightarrow f(x) =  - \frac{1}{{{x^2}}} + \frac{{ - \frac{1}{3}x + \frac{1}{3}}}{{{x^2} + x + 1}} + \frac{{\frac{1}{3}}}{{x - 1}}$
Vậy : $I = \int\limits_2^3 {\left( { - \frac{1}{{{x^2}}} + \frac{{ - \frac{1}{3}x + \frac{1}{3}}}{{{x^2} + x + 1}} + \frac{{\frac{1}{3}}}{{x - 1}}} \right)dx}  = \int\limits_2^3 {\left( { - \frac{1}{{{x^2}}} - \frac{1}{3}\left( {\frac{{x - 1}}{{{x^2} + x + 1}}} \right) + \frac{1}{3}\frac{1}{{\left( {x - 1} \right)}}} \right)dx}$
$= \left( {\frac{1}{x} - \frac{1}{6}\ln \left| {{x^2} + x + 1} \right| + \frac{1}{3}\ln \left| {x - 1} \right|} \right)\left| {\begin{array}{*{20}{c}}   3 \\   2 \end{array} - \int\limits_2^3 {\frac{{dx}}{{{{\left( {x + \frac{1}{2}} \right)}^2} + {{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}}} } \right.\\ = \left( {\frac{1}{x} + \frac{1}{6}\ln \frac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} + x + 1}} + \frac{1}{{\sqrt 3 }}{\text{arctan}}\frac{{{\text{2x + 1}}}}{{\sqrt 3 }}} \right)\left| {\begin{array}{*{20}{c}}   3 \\   2 \end{array}} \right.$$= \frac{1}{6} + \frac{1}{{\sqrt 3 }}\left( {{\text{arctan}}\frac{{\text{7}}}{{\sqrt {\text{3}} }} - {\text{arctan}}\frac{{\text{5}}}{{\sqrt {\text{3}} }}} \right)$
b. $\int\limits_0^1 {\frac{{{x^7}dx}}{{{{\left( {1 + {x^4}} \right)}^2}}}}  = \frac{1}{3}\int\limits_0^1 {\frac{{{x^4}}}{{{{\left( {1 + {x^4}} \right)}^2}}}3{x^3}dx\quad \left( 1 \right)}$.
Đặt : $t = 1 + {x^4} \Rightarrow \left\{ \begin{array}   dt = 3{x^3}dx,x = 0 \to t = 1;x = 1 \to t = 2  \\   f(x)dx = \frac{1}{3}\left( {\frac{{t - 1}}{{{t^2}}}} \right)dt = \frac{1}{3}\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right)dt  \\ \end{array}  \right.$
Vậy : $I = \int\limits_0^2 {\frac{1}{3}\left( {\frac{1}{t} - \frac{1}{{{t^2}}}} \right)dt = \frac{1}{3}\left( {\ln \left| t \right| + \frac{1}{t}} \right)\left| {\begin{array}{*{20}{c}}   2 \\   1 \end{array} = \frac{1}{3}\left( {\ln 2 - \frac{1}{2}} \right)} \right.}$
c. $\int\limits_0^1 {\frac{{{x^3} - 2x}}{{{{\left( {{x^2} + 1} \right)}^2}}}dx}  = \frac{1}{2}\int\limits_0^1 {\frac{{\left( {{x^2} - 2} \right)}}{{{{\left( {{x^2} + 1} \right)}^2}}}2xdx} \quad \left( 1 \right)$
Đặt : $t = 1 + {x^2} \Leftrightarrow {x^2} - 2 = t - 3 \Rightarrow \left\{ \begin{array}   dt = 2xdx;x = 0 \to t = 1;x = 1 \to t = 2  \\   f(x)dx = \frac{1}{2}\left( {\frac{{t - 3}}{{{t^2}}}} \right)dt = \frac{1}{2}\left( {\frac{1}{t} - \frac{3}{{{t^2}}}} \right)dt  \\ \end{array}  \right.$
Vậy : $I = \int\limits_1^2 {\frac{1}{2}\left( {\frac{1}{t} - \frac{3}{{{t^2}}}} \right)dt = \frac{1}{2}\left( {\ln \left| t \right| + \frac{3}{t}} \right)\left| {\begin{array}{*{20}{c}}   2 \\   1 \end{array} = \frac{1}{2}\left( {\ln 2 - \frac{3}{2}} \right)} \right.}$
d. $\int\limits_1^2 {\frac{{\sqrt {1 + {x^3}} }}{{{x^4}}}dx}  = \int\limits_1^2 {\frac{{\sqrt {1 + {x^3}} }}{{{x^6}}}{x^2}dx} \quad \left( 1 \right)$.
Đặt : $t = \sqrt {1 + {x^3}}  \leftrightarrow {t^2} = 1 + {x^3} \leftrightarrow \left\{ \begin{array}   2tdt = 3{x^2}dx;x = 1 \to t = \sqrt 2 ,x = 2 \to t = 3  \\   f(x)dx = \frac{1}{3}\frac{{\sqrt {1 + {x^3}} }}{{{x^6}}}3{x^2}dx = \frac{1}{3}\frac{t}{{{{\left( {{t^2} - 1} \right)}^2}}}2tdt = \frac{2}{3}\frac{{{t^2}}}{{{{\left( {{t^2} - 1} \right)}^2}}}dt  \\ \end{array}  \right.$
Vậy : $I = \frac{2}{3}\int\limits_{\sqrt 2 }^3 {{{\left( {\frac{1}{{t + 1}} + \frac{1}{2}\left( {\frac{1}{{t - 1}} - \frac{1}{{t + 1}}} \right)} \right)}^2}dt = \frac{2}{3}\left[ {\int\limits_{\sqrt 2 }^3 {\frac{1}{4}{{\left( {\frac{1}{{t + 1}} - \frac{1}{{t - 1}}} \right)}^2}} } \right]}$
$= \frac{1}{6}\int\limits_{\sqrt 2 }^3 {\left( {\frac{1}{{{{\left( {t + 1} \right)}^2}}} + \frac{1}{{{{\left( {t - 1} \right)}^2}}} - \left( {\frac{1}{{t - 1}} - \frac{1}{{t + 1}}} \right)} \right)dt}$
$= \frac{1}{6}\left[ { - \frac{1}{{t + 1}} - \frac{1}{{t - 1}} - \ln \left| {\frac{{t - 1}}{{t + 1}}} \right|} \right]\left| {\begin{array}{*{20}{c}}   3 \\   {\sqrt 2 } \end{array} = \frac{1}{6}\left( {\frac{{ - 2t}}{{\left( {{t^2} - 1} \right)}} - \ln \left| {\frac{{t - 1}}{{t + 1}}} \right|} \right)\left| {\begin{array}{*{20}{c}}   3 \\   {\sqrt 2 } \end{array}} \right.} \right. = \frac{{8\sqrt 2  - 3}}{{24}} + \frac{1}{3}\ln \left( {2\sqrt 2  - 2} \right)$

Ví dụ 5.

Tính các tích phân sau :
a. $\int\limits_{\sqrt 7 }^4 {\frac{{dx}}{{x\sqrt {{x^2} + 9} }}}$                    b. $\int\limits_0^1 {\frac{{\left( {{x^2} - x} \right)dx}}{{\sqrt {{x^2} + 1} }}}$
 c. $\int\limits_0^{\sqrt 3 } {\frac{{{x^5} - 2{x^3}}}{{\sqrt {{x^2} + 1} }}dx}$                    d. $\int\limits_0^1 {\sqrt {{{\left( {1 - {x^2}} \right)}^3}} dx}$
Giải
a. $\int\limits_{\sqrt 7 }^4 {\frac{{dx}}{{x\sqrt {{x^2} + 9} }}}  = \int\limits_{\sqrt 7 }^4 {\frac{{xdx}}{{{x^2}\sqrt {{x^2} + 9} }}} \quad \left( 1 \right)$.
Đặt : $t = \sqrt {{x^2} + 9}  \Rightarrow \left\{ \begin{array}   {t^2} = {x^2} + 9 \leftrightarrow tdt = xdx,{x^2} = {t^2} - 9  \\   x = \sqrt 7  \to t = 4,x = 4 \to t = 5  \\ \end{array}  \right.$. Do đó : $I = \int\limits_4^5 {\frac{{dt}}{{t\left( {{t^2} - 9} \right)}} = } \int\limits_4^5 {\frac{{dt}}{{t\left( {t - 3} \right)\left( {t + 3} \right)}}}$
Ta có : $f(t) = \frac{1}{{t\left( {t - 3} \right)\left( {t + 3} \right)}} = \frac{A}{t} + \frac{B}{{t - 3}} + \frac{C}{{t + 3}} = \frac{{A\left( {{t^2} - 9} \right) + Bt\left( {t + 3} \right) + C\left( {t - 3} \right)t}}{{t\left( {{t^2} - 9} \right)}}$
Đồng nhất hệ số hai tử số bằng cách thay lần lượt các nghiệm vào hai tử số ta có :
- Với x=0 : -9A=1 $\to A =  - \frac{1}{9}$
- Với x=-3 : 9C=1 $\to C = \frac{1}{9}$
- Với x=3 : 9B=1 $\to B = \frac{1}{9}$
Vậy : $I = \frac{1}{9}\left[ {\int\limits_4^5 {\left( { - \frac{1}{t} + \frac{1}{{t - 3}} + \frac{1}{{t + 3}}} \right)dt} } \right] = \frac{1}{9}\left[ {\ln \left( {{t^2} - 9} \right) - \ln t} \right]\left| {\begin{array}{*{20}{c}}   5 \\   4 \end{array} = \frac{1}{9}\ln \frac{{{t^2} - 9}}{t}\left| {\begin{array}{*{20}{c}}   5 \\   4 \end{array} = \frac{1}{9}\ln \frac{{144}}{{35}}} \right.} \right.$
Chú ý : Nếu theo phương pháp chung thì đặt : $x = 3\sin t \to dx = 3\cos tdt$.
Khi : $\left\{ \begin{array}   x = \sqrt 7  \to \sqrt 7  = 3\sin t \leftrightarrow \sin t = \frac{{\sqrt 7 }}{3}  \\   x = 4 \to 4 = 3\sin t \leftrightarrow \sin t = \frac{4}{3} > 1  \\ \end{array}  \right.$. Như vậy ta không sử dụng được phương pháp này được .
b. $\int\limits_0^1 {\frac{{\left( {{x^2} - x} \right)dx}}{{\sqrt {{x^2} + 1} }}}  = \int\limits_0^1 {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}dx - \int\limits_0^1 {\frac{x}{{\sqrt {{x^2} + 1} }}dx}  = J - K\quad \left( 1 \right)}$
* Để tính J :
Đặt : $x = \tan t \Rightarrow \left\{ \begin{array}   dx = \frac{1}{{c{\text{o}}{{\text{s}}^2}t}}dt,x = 0 \to t = 0;x = 1 \to t = \frac{\pi }{4}  \\   f(x)dx = \frac{{{{\tan }^2}t.\frac{1}{{c{\text{o}}{{\text{s}}^2}t}}dt}}{{\sqrt {1 + {{\tan }^2}t} }} = \frac{{{{\tan }^2}t}}{{c{\text{ost}}}}dt  \\ \end{array}  \right.$. Tính tích phân này không đơn giản , vì vậy ta phải có cách khác .
- Từ : $g(x) = \frac{{{x^2}}}{{\sqrt {{x^2} + 1} }} = \frac{{{x^2} + 1 - 1}}{{\sqrt {{x^2} + 1} }} = \sqrt {{x^2} + 1}  - \frac{1}{{\sqrt {{x^2} + 1} }} \Rightarrow \int\limits_0^1 {g(x)dx = \int\limits_0^1 {\sqrt {{x^2} + 1} dx - \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}dx} } }$
- Hai tích phân này đều tính được .
+/ Tính : $E = \int\limits_0^1 {\sqrt {{x^2} + 1} dx = } x\sqrt {{x^2} + 1} \left| {\begin{array}{*{20}{c}}   1 \\   0 \end{array} - \int\limits_0^1 {\frac{{{x^2}}}{{\sqrt {{x^2} + 1} }}dx = } } \right.\sqrt 2  - \left( {\int\limits_0^1 {\sqrt {{x^2} + 1} } dx - \int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}dx} } \right)$
$= \sqrt 2  - E + \ln \left| {x + \sqrt {{x^2} + 1} } \right|\left| {\begin{array}{*{20}{c}}   1 \\   0 \end{array}} \right. \Rightarrow 2E = \sqrt 2  + \ln \left( {1 + \sqrt 2 } \right) \Leftrightarrow E = \frac{{\sqrt 2 }}{2} + \frac{1}{2}\ln \left( {1 + \sqrt 2 } \right)$
* Tính K=$\int\limits_0^1 {\frac{x}{{\sqrt {{x^2} + 1} }}dx = \sqrt {{x^2} + 1} \left| {\begin{array}{*{20}{c}}   1 \\   0 \end{array} = \sqrt 2  - 1} \right.}$; $\int\limits_0^1 {\frac{1}{{\sqrt {{x^2} + 1} }}dx = \ln \left| {x + \sqrt {{x^2} + 1} } \right|\left| {\begin{array}{*{20}{c}}   1 \\   0 \end{array} = \ln \left( {1 + \sqrt 2 } \right)} \right.}$
Do vậy : I=$\frac{{\sqrt 2 }}{2} + \frac{1}{2}\ln \left( {1 + \sqrt 2 } \right) + \ln \left( {1 + \sqrt 2 } \right) = \frac{{\sqrt 2 }}{2} + \frac{3}{2}\ln \left( {1 + \sqrt 2 } \right)$
c. $\int\limits_0^{\sqrt 3 } {\frac{{{x^5} - 2{x^3}}}{{\sqrt {{x^2} + 1} }}dx}  = \int\limits_0^{\sqrt 3 } {\frac{{{x^5}}}{{\sqrt {{x^2} + 1} }}dx - 2\int\limits_0^{\sqrt 3 } {\frac{{{x^3}}}{{\sqrt {{x^2} + 1} }}dx = J - K\left( 1 \right)} }$
- Tính J: Đặt $t = \sqrt {{x^2} + 1}  \Rightarrow \left\{ \begin{array}   {x^2} = {t^2} - 1;xdx = tdt;x = 0 \to t = 1,x = \sqrt 3  \to t = 2  \\   f(x)dx = \frac{{{x^4}xdx}}{{\sqrt {{x^2} + 1} }} = \frac{{{{\left( {{t^2} - 1} \right)}^2}tdt}}{t} = \left( {{t^4} - 2{t^2} + 1} \right)dt  \\ \end{array}  \right.$
Suy ra : J=$\int\limits_1^2 {\left( {{t^4} - 2{t^2} + 1} \right)dt = \left( {\frac{1}{5}{t^5} - \frac{2}{3}{t^3} + t} \right)\left| {\begin{array}{*{20}{c}}   2 \\   1 \end{array} = \frac{{38}}{{15}}} \right.}$
- Tính K: Đặt $t = \sqrt {{x^2} + 1}  \Rightarrow \left\{ \begin{array}   {x^2} = {t^2} - 1;xdx = tdt;x = 0 \to t = 1,x = \sqrt 3  \to t = 2  \\   f(x)dx = \frac{{{x^2}xdx}}{{\sqrt {{x^2} + 1} }} = \frac{{\left( {{t^2} - 1} \right)tdt}}{t} = \left( {{t^2} - 1} \right)dt  \\ \end{array}  \right.$
Suy ra : K= $\int\limits_1^2 {\left( {{t^2} - 1} \right)dt = \left( {\frac{1}{3}{t^3} - t} \right)\left| {\begin{array}{*{20}{c}}   2 \\   1 \end{array} = \frac{4}{3}} \right.}$
Vậy : I=$\frac{{28}}{{15}} + \frac{4}{3} = \frac{{48}}{{15}} = \frac{{16}}{5}$
d. $\int\limits_0^1 {\sqrt {{{\left( {1 - {x^2}} \right)}^3}} dx}$. Đặt : $x = \sin t \to \left\{ \begin{array}   dx = c{\text{ostdt}}{\text{. x = 0}} \to {\text{t = 0;x = 1}} \to {\text{t = }}\frac{\pi }{2}  \\   f(x)dx = \sqrt {{{\left( {1 - {x^2}} \right)}^3}} dx = \sqrt {c{\text{o}}{{\text{s}}^{\text{6}}}t} c{\text{ostdt = co}}{{\text{s}}^{\text{4}}}tdt  \\ \end{array}  \right.$
Do đó I=$\int\limits_0^{\frac{\pi }{2}} {{{\left( {\frac{{1 - c{\text{os2t}}}}{2}} \right)}^2}dt = \frac{1}{4}\int\limits_0^{\frac{\pi }{2}} {\left( {1 - 2\cos 2t + \frac{{1 + c{\text{os4t}}}}{2}} \right)dt = \int\limits_0^{\frac{\pi }{2}} {\left( {\frac{3}{4} - \frac{1}{2}c{\text{os2t + }}\frac{1}{{\text{8}}}c{\text{os4t}}} \right)dt} } }$
            $= \left( {\frac{3}{4}t - \frac{1}{4}\sin 2t + \frac{1}{{32}}\sin 4t} \right)\left| {\begin{array}{*{20}{c}}   {\frac{\pi }{2}} \\   0 \end{array} = \frac{{3\pi }}{8}} \right.$