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,

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

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

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

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

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

 

Hence, this is the answer.

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