Bài 106233

Question

Tính các tích phân sau:
a)$ \int\limits_{}^{}\frac{1+e^{3x}}{1+e^{x}}dx$               b)$ \int\limits_{}^{}\frac{x^{4}}{x^{2}-1}dx$
c)$ \int\limits_{}^{}\tan ^{8}xdx$                                                      d) $\int\limits_{}^{}\left ( \cot x- \tan x\right )^{2}dx$
e)$ \int\limits_{}^{}\sqrt{e^{2x}+e^{-2x}+2}dx$                                  f)$\int\limits_{}^{}\frac{dx}{3x^{2}-2} $

score 0 · Answer 1

a) $ \frac{1+e^{3x}}{1+e^{x}}=1-e^{x}+e^{2x}$
$\Rightarrow \int\limits \frac{1+e^{3x}}{1+e^x}dx $
$=\int\limits (1-e^x+e^{2x})dx= x-e^x+\frac{1}{2}e^{2x}+C $
b)$ \frac{x^{4}}{x^{2}-1}=\frac{\left ( x^{4}-1 \right )+1}{x^{2}-1}=x^{2}+1+\frac{1}{2}\left ( \frac{1}{x-1}-\frac{1}{x+1} \right )$
$\Rightarrow \int\limits \frac{x^4}{x^2-1}dx = \int\limits \left[ {x^2+1+\frac{1}{2}   \left ( \frac{1}{x-1} -\frac{1}{x+1} \right ) } \right] dx$
$=\frac{x^3}{3}+x+\frac{1}{2} \ln |\frac{x-1}{x+1} | +C $
c)$ \tan ^{8}x=\left (\tan ^{8}x+\tan ^{6}x \right )-\left ( \tan ^{6}x+\tan ^{4}x \right )+\left ( \tan ^{4}x+\tan ^{2}x \right )$
$-\left ( \tan ^{2}x+1 \right )+1$
$ =\tan ^{6}x\left ( \tan ^{2}x+1 \right )-\tan ^{4}x\left ( \tan ^{2}x+1 \right )+\tan ^{2}x\left ( \tan ^{2}x+1 \right )$
$-\left ( \tan ^{2}x+1 \right )+1$
$=\tan ^{6}x\left ( \tan x \right )^{'}-\tan ^{4}x\left ( \tan x \right )^{'}+\tan ^{2}x\left ( \tan x \right )^{'}-\left ( \tan x \right )^{'}+1$
$\Rightarrow \int\limits \tan ^8xdx=\int\limits   \left ( \tan ^6x-\tan ^4x+\tan ^2x-1 \right ) d(\tan x)+ \int\limits dx$
$= \frac{\tan ^7x}{7}- \frac{\tan ^5x}{5}+\frac{\tan ^3x}{3}-\tan x+C   $
d)$\left ( \cot x-\tan x \right )^{2}=\cot ^{2}x+\tan ^{2}x-2=\left ( 1+\cot ^{2}x\right )+\left ( 1+\tan ^{2}x\right )-4 $
$\Rightarrow \int\limits   \left ( \cot x-\tan x \right )^2dx = \int\limits \left[ { \left ( 1+ \cot ^2x \right ) +   \left ( 1+\tan ^2x \right )-4 } \right]dx $
$=- \int\limits d   \left ( \cot x \right )+ \int\limits d (\tan x) - \int\limits 4 dx$
$=- \cot x+\tan x-4x+C$
e)$e^{2x}+e^{-2x}+2=\left (e^{x}+e^{-x} \right )^{2} $
$\Rightarrow \int\limits \sqrt{e^{2x}+e^{-2x}+2}dx=\int\limits   \left ( e^x+e^{-x} \right ) dx= e^x-e^{-x}+C $
f)$ \frac{1}{3x^2-2}=\frac{1}{3}\frac{1}{x^{2}-\frac{2}{3}}=\frac{1}{2\sqrt{6}}\left ( \frac{1}{x-\sqrt{\frac{2}{3}}}-\frac{1}{x+\sqrt{\frac{2}{3}}} \right )$
$\Rightarrow \int\limits \frac{dx}{3x^2-2}=\frac{1}{2 \sqrt{6} } \int\limits \left[ {\frac{1}{x- \sqrt{\frac{2}{3} } }- \frac{1}{x+ \sqrt{\frac{2}{3} } }} \right]   $
$= \frac{1}{2 \sqrt{6} } \ln |\frac{x- \sqrt{\frac{2}{3} } }{x+ \sqrt{\frac{2}{3} } } |+C$
$=\frac{1}{2 \sqrt{6} }\ln |\frac{\sqrt{3}x-\sqrt{2}}{\sqrt{3}x+\sqrt{2}}   |+C$

Bài 106233

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Bài 106233

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