Mathematics

# Evaluate  :  $\displaystyle \int \dfrac{\tan^{ - 1}x}{1+x^2}\ dx$

##### SOLUTION
We have,
$\displaystyle I=\int \dfrac{\tan^{ - 1}x}{1+x^2}\ dx$

Let $t=\tan^{-1}x$

$dt=\dfrac{1}{1+x^2}dx$

Therefore,
$\displaystyle I=\int t\ dt$

$I=\dfrac{t^2}{2}+C$

$I=\dfrac{(\tan^{-1}x)^2}{2}+C$

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

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