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Question Number 54209 by cesar.marval.larez@gmail.com last updated on 31/Jan/19

Answered by Joel578 last updated on 31/Jan/19

$$I=\int \frac{{\tan }^{-1}x}{1+x^2}dx$$$$u={\tan }^{-1}x\to du=\frac{dx}{1+x^2}$$$$dv=\frac{dx}{1+x^2}\to v={\tan }^{-1}x$$$$I={\left({\tan }^{-1}x\right)}^2-\int \frac{{\tan }^{-1}x}{1+x^2}dx$$$$={\left({\tan }^{-1}x\right)}^2-I$$$$2I={\left({\tan }^{-1}x\right)}^2$$$$I=\frac{1}{2}{\left({\tan }^{-1}x\right)}^2+C$$

Answered by Prithwish sen last updated on 31/Jan/19

$$$$$$\mathrm{let},{\tan }^{-1}\mathrm{x}=\mathrm{t}$$$$\mathrm{then}\frac{\mathrm{dx}}{1+x^2}=dt$$$$\therefore \int \frac{{\tan }^{-1}\mathrm{x}}{1+{\mathrm{x}}^2}\mathrm{dx}=\int \mathrm{t}\mathrm{dt}$$$$=\frac{\mathrm{t}2}{2}+\mathrm{c}$$$$=\frac{1}{2}{\left({\tan }^{-1}\mathrm{x}\right)}^2+\mathrm{c}$$

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