Mathematics

Integrate with respect to $$'x'$$:
$$x\tan ^{ -1 }{ x } $$


SOLUTION
$$\displaystyle\int x{\text{tan}^{-1}x}dx=\text{tan}^{-1}x\displaystyle\int x{dx}-\displaystyle\int \dfrac{d(\text{tan}^{-1}x)}{dx}\bigg(\displaystyle\int x{dx}\bigg)dx$$
$$=\dfrac{x^{2}\text{tan}^{-1}x}{2}-\dfrac{1}{2}\displaystyle\int \dfrac{x^{2}+1-1}{1+x^{2}}dx\\=\dfrac{x^{2}\text{tan}^{-1}x}{2}-\dfrac{x}{2}+\dfrac{\text{tan}^{-1}x}{2}$$
$$=\dfrac{(x^{2}+1)\text{tan}^{-1}x-x}{2}$$
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Subjective Medium Published on 17th 09, 2020
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