tích phân hàm hữu tỉ

Question

tính tích phân a, $I=\int\limits_{0}^{1}\frac{x^{3}-4x^{2}+11x-1}{x^{2}-4x+8}dx$
b, $I=\int\limits_{-1}^{0}\frac{x}{x^{3}-5x+2}dx$
c,$I=\int\limits_{0}^{1}\frac{3x^{2}}{x^{2}+4x+4}dx$
d,$I=\int\limits_{1}^{\sqrt{2}}\frac{dx}{2x+x^{3}}$
e, $I=\int\limits_{1}^{2} \frac{dx}{x+x^{5}}dx$

em mong được học hỏi nhiều hơn ! thanhk nhiều naz

score 7 · Answer 1

Bình chọn tăng 7 Bình chọn giảm

e,
Ta có:
$I=\int\limits_{1}^{2} \frac{dx}{x+x^{5}}$
    $=\int\limits_1^2\left(\frac{1}{x}-\frac{x^3}{1+x^4}\right)dx$
    $=\int\limits_1^2\frac{dx}{x}-\frac{1}{4}\int\limits_1^2\frac{d(x^4+1)}{x^4+1}$
    $=\ln x \left|\begin{array}{l}2\\1\end{array}\right.-\frac{1}{4}\ln(1+x^4)\left|\begin{array}{l}2\\1\end{array}\right. $
    $=\frac{1}{4}\ln\frac{32}{17}$

Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! – khangnguyenthanh 15-01-13 08:34 PM

score 7 · Answer 2

Bình chọn tăng 7 Bình chọn giảm

d,
Ta có:
$I=\int\limits_{1}^{\sqrt{2}}\frac{dx}{2x+x^{3}}$
    $=\int\limits_1^{\sqrt2}\left(\frac{1}{2x}-\frac{x}{2(x^2+2)}\right)dx$
    $=\frac{1}{2}\int\limits_1^{\sqrt2}\frac{dx}{x}-\frac{1}{4}\int\limits_1^{\sqrt2}\frac{d(x^2+2)}{x^2+2}$
    $=\frac{1}{2}\ln x \left|\begin{array}{l}\sqrt2\\1\end{array}\right.-\frac{1}{4}\ln(x^2+2)\left|\begin{array}{l}\sqrt2\\1\end{array}\right. $
    $=\frac{1}{4}\ln\frac{3}{2}$

Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! – khangnguyenthanh 15-01-13 08:34 PM

score 7 · Answer 3

Bình chọn tăng 7 Bình chọn giảm

c,
Ta có:
$I=\int\limits_{0}^{1}\frac{3x^{2}}{x^{2}+4x+4}dx$
    $=\int\limits_0^1\left(3-\frac{12}{x+2}+\frac{12}{x^2+4x+4}\right)dx$
    $=\int\limits_0^13dx-12\int\limits_0^1\frac{d(x+2)}{x+2}+12\int\limits_0^1\frac{d(x+2)}{(x+2)^2}$
    $=3x \left|\begin{array}{l}1\\0\end{array}\right. -12\ln(x+2)\left|\begin{array}{l}1\\0\end{array}\right. -\frac{12}{x+2} \left|\begin{array}{l}1\\0\end{array}\right. $
    $=5+12\ln2-12\ln3$

Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! – khangnguyenthanh 15-01-13 08:34 PM

score 7 · Answer 4

Bình chọn tăng 7 Bình chọn giảm

b,
Ta có:
$I=\int\limits_{-1}^{0}\frac{x}{x^{3}-5x+2}dx$
    $=\int\limits_{-1}^0\left(\frac{2}{7(x-2)}-\frac{2x+1}{7(x^2+2x-1)}\right)dx$
    $=\int\limits_{-1}^0\left(\frac{2}{7(x-2)}-\frac{1+2\sqrt2}{14\sqrt2(x+1+\sqrt2)}+\frac{2\sqrt2-1}{14\sqrt2(x+1-\sqrt2)}\right)dx$
    $=\frac{2}{7}\int\limits_{-1}^0\frac{d(x-2)}{x-2}-\frac{1+2\sqrt2}{14\sqrt2}\int\limits_{-1}^0\frac{d(x+1+\sqrt2)}{x+1+\sqrt2}+\frac{2\sqrt2-1}{14\sqrt2}\int\limits_{-1}^0\frac{d(x+1-\sqrt2)}{x+1-\sqrt2}$
    $=\frac{2}{7}\ln|x-2|\left|\begin{array}{l}0\\-1\end{array}\right.-\frac{1+2\sqrt2}{14\sqrt2}\ln|x+1+\sqrt2|\left|\begin{array}{l}0\\-1\end{array}\right.+\frac{2\sqrt2-1}{14\sqrt2}\ln|x+1-\sqrt2|\left|\begin{array}{l}0\\-1\end{array}\right.$
    $=-\frac{2}{7}\ln3+\frac{3}{7}\ln2-\frac{\sqrt{2}}{14}\ln(1+\sqrt2)$

Nếu thấy lời giải đúng thì bạn vui lòng đánh dấu vào hình chữ V dưới phần vote để xác nhận nhá. Thanks! – khangnguyenthanh 15-01-13 08:34 PM

score 5 · Answer 5

a,
Ta có:
$I=\int\limits_{0}^{1}\frac{x^{3}-4x^{2}+11x-1}{x^{2}-4x+8}dx$
   $=\int\limits_0^1\left(x+\frac{3(2x-4)}{2(x^2-4x+8)}+\frac{5}{x^2-4x+8}\right)dx$
   $=\int\limits_0^1xdx+\frac{3}{2}\int\limits_0^1\frac{d(x^2-4x+8)}{x^2-4x+8}+5\int\limits_0^1\frac{dx}{(x-2)^2+4}$
   $=\frac{x^2}{2}\left|\begin{array}{l}1\\0\end{array}\right.+\frac{3}{2}\ln(x^2-4x+8)\left|\begin{array}{l}1\\0\end{array}\right.+5\int\limits_0^1\frac{dx}{(x-2)^2+4}$
   $=\frac{1}{2}+\frac{3}{2}\ln\frac{5}{8}+5\int\limits_0^1\frac{dx}{(x-2)^2+4}$
Đặt $x-2=2\tan t, t\in[-\frac{\pi}{2};\frac{\pi}{2}]$
Suy ra: $dx=2d(\tan t)=2(\tan^2t+1)dt$
Đổi cận: $x=0\Rightarrow t=-\frac{\pi}{4}$
                $x=1\Rightarrow t=\arctan(-\frac{1}{2})$
Từ đó suy ra:
$\int\limits_0^1\frac{dx}{(x-2)^2+4}=\int\limits_{-\pi/4}^{\arctan(-1/2)}\frac{2(\tan^2t+1)dt}{4\tan^2t+4}$
                    $=\frac{1}{2}\int\limits_{-\pi/4}^{\arctan(-1/2)}dt$
                    $=\frac{1}{2}t \left|\begin{array}{l}\arctan(-\frac{1}{2})\\-\frac{\pi}{4}\end{array}\right.=\frac{\pi}{8}-\frac{1}{2}\arctan(\frac{1}{2})$
Suy ra: $I=\frac{1}{2}+\frac{3}{2}\ln\frac{5}{8}+\frac{5\pi}{8}-\frac{5}{2}\arctan(\frac{1}{2})$