Mathematics

# Integrate:$\displaystyle \int \dfrac{\sqrt{1+x^{2}}}{x}dx$

##### SOLUTION
$\int \dfrac{\sqrt{1+x^2}}{x}dx$

substitute  $u=\sqrt{1+x^2}\rightarrow dx=\dfrac{du}{u^2-1}$

$=\int \dfrac{u^2}{u^2-1}du$

$=-\int \dfrac{u^2}{1-u^2}du$

$=-\int \left ( \dfrac{1}{1-u^2}-1 \right )du$

$=-\left ( \dfrac{\ln |u+1|}{2} -\dfrac{\ln |u-1|}{2}-u\right )$

$=-\left ( \dfrac{\ln |\sqrt{1+x^2}+1|}{2} -\dfrac{\ln |\sqrt{1+x^2}-1|}{2}-\sqrt{1+x^2}\right )+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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