Mathematics

# Simplify:$\int {x{{\tan }^{ - 1}}xdx}$

##### SOLUTION

Consider the given integral.

$I=\int{x{{\tan }^{-1}}xdx}$

We know that

$\int{uvdx=u\int{vdx-\int{\left( \dfrac{d}{dx}\left( u \right)\int{vdx} \right)}}}dx$

Therefore,

$I={{\tan }^{-1}}x\left( \dfrac{{{x}^{2}}}{2} \right)-\int{\dfrac{1}{1+{{x}^{2}}}\times \left( \dfrac{{{x}^{2}}}{2} \right)}dx$

$I=\dfrac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\dfrac{1}{2}\int{\dfrac{{{x}^{2}}}{1+{{x}^{2}}}}dx$

$I=\dfrac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\dfrac{1}{2}\int{\dfrac{{{x}^{2}}+1-1}{1+{{x}^{2}}}}dx$

$I=\dfrac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\dfrac{1}{2}\left[ \int{1}dx-\int{\dfrac{1}{1+{{x}^{2}}}}dx \right]$

$I=\dfrac{{{x}^{2}}}{2}{{\tan }^{-1}}x-\dfrac{1}{2}\left[ x-{{\tan }^{-1}}x \right]+C$

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

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