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Question Number 130214 by Lordose last updated on 23/Jan/21

$$\int \frac{{\mathrm{u}}^2}{{(1+{\mathrm{u}}^2)}^2}\mathrm{du}$$

Answered by liberty last updated on 23/Jan/21

$$\mathrm{J}=\int \frac{({\mathrm{u}}^2+1)-1}{{(1+{\mathrm{u}}^2)}^2}\mathrm{du}=\int \frac{\mathrm{du}}{1+{\mathrm{u}}^2}-\int \frac{\mathrm{du}}{{(1+{\mathrm{u}}^2)}^2}$$$${\mathrm{J}}_1=\int \frac{\mathrm{du}}{1+{\mathrm{u}}^2}={\tan }^{-1}\left(\mathrm{u}\right)+{\mathrm{c}}_1$$$${\mathrm{J}}_2=\int \frac{\mathrm{du}}{{(1+{\mathrm{u}}^2)}^2};\mathrm{u}=\tan \mathrm{q}$$$${\mathrm{J}}_2=\int \frac{\sec {}^2\mathrm{q}\mathrm{dq}}{\sec {}^4\mathrm{q}}=\int \cos {}^2\mathrm{q}\mathrm{dq}$$$${\mathrm{J}}_2=\frac{1}{2}\mathrm{q}+\frac{1}{4}\sin 2\mathrm{q}+{\mathrm{c}}_2$$$${\mathrm{J}}_2=\frac{{\tan }^{-1}\left(\mathrm{u}\right)}{2}+\frac{\mathrm{u}}{2(1+{\mathrm{u}}^2)}+{\mathrm{c}}_2$$$$\therefore \mathrm{J}=\frac{1}{2}{\tan }^{-1}\left(\mathrm{u}\right)-\frac{1}{2}\frac{\mathrm{u}}{{(1+\mathrm{u})}^2}+\mathrm{C}$$

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