giup minh voi

$K = \int\limits_1^2 {\frac{{dx}}{{x\sqrt {{x^3} + 1} }}}  = \int\limits_1^2 {\frac{{{x^2}dx}}{{{x^3}\sqrt {{x^3} + 1} }}} $

Đặt: \[t = \sqrt {{x^3} + 1}  \Leftrightarrow {t^2} = {x^3} + 1 \Leftrightarrow \frac{{2tdt}}{3} = {x^2}dx\]\[x = 2 \Rightarrow t = 3\]\[x = 1 \Rightarrow t = \sqrt 2 \]\[ \Rightarrow K = \frac{2}{3}\int\limits_{\sqrt 2 }^3 {\frac{{tdt}}{{({t^2} - 1)t}}}  = \frac{2}{3}\int\limits_{\sqrt 2 }^3 {\frac{{dt}}{{{t^2} - 1}}}  = \frac{1}{3}\int\limits_{\sqrt 2 }^3 {(\frac{1}{{t - 1}}}  - \frac{1}{{t + 1}})dt = \frac{1}{3}\left( {\ln \left| {t - 1} \right| - \ln \left| {t + 1} \right|} \right) = \frac{1}{3}\ln \frac{{{{(\sqrt 2  + 1)}^2}}}{2}\]

$J = \int\limits_{ - 1}^0 {\frac{{dx}}{{{x^2} + 2x + 4}}}  = \int\limits_{ - 1}^0 {\frac{{dx}}{{{{(x + 1)}^2} + 3}}} $

Đặt \[x + 1 = \sqrt 3 \tan t,t \in \left( { - \frac{\pi }{2};\frac{\pi }{2}} \right)\]

\[dx = \sqrt 3 ({\tan ^2}t + 1)dt\]\[x = 0 \Rightarrow t = \frac{\pi }{6}\]\[x =  - 1 \Rightarrow t = 0\]

\[ \Rightarrow J = \int\limits_0^{\frac{\pi }{6}} {\frac{{\sqrt 3 ({{\tan }^2}t + 1)dt}}{{3{{\tan }^2}t + 3}}}  = \frac{{\sqrt 3 }}{3}\int\limits_0^{\frac{\pi }{6}} {dt}  = \frac{{\sqrt 3 }}{3}.\frac{\pi }{6} = \frac{{\pi \sqrt 3 }}{{18}}\]

\[\frac{{{x^2}}}{{{{(x + 2)}^2}}} = \frac{{({x^2} + 4x + 4) - 4(x + 2) + 4}}{{{{(x + 2)}^2}}} = 1 - \frac{4}{{x + 2}} + \frac{4}{{{{(x + 2)}^2}}}\]

\[ \Rightarrow I = \int\limits_0^2 {{e^x}dx - 4} \int\limits_0^2 {\frac{{{e^x}}}{{x + 2}}dx + 4} \int\limits_0^2 {\frac{{{e^x}}}{{{{(x + 2)}^2}}}dx} \]

Ta tính tích phân $\int\limits_0^2 {\frac{{{e^x}}}{{x + 2}}dx}$:

Đặt: \[u = \frac{1}{{x + 2}} \Rightarrow du =  - \frac{{dx}}{{{{(x + 2)}^2}}}\]

\[dv = {e^x}dx \Leftarrow v = {e^x}\]

$ \Rightarrow I = {e^2} - 1 - 4\left( {\frac{{{e^x}}}{{x + 2}}\begin{cases}2 \\ 0 \end{cases} + \int\limits_0^2 {\frac{{{e^x}}}{{{{(x + 2)}^2}}}dx} } \right) + 4\int\limits_0^2 {\frac{{{e^x}}}{{{{(x + 2)}^2}}}dx}$

$I = {e^2} - 1 - 4(\frac{{{e^2}}}{4} - \frac{1}{2}) = 1$