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Question Number 176187 by gloriousman last updated on 14/Sep/22

∫(dy/(tan^(−1) y(1+y^2 )))

$$\int\frac{\mathrm{dy}}{\mathrm{tan}^{−\mathrm{1}} \mathrm{y}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)} \\ $$

Answered by BaliramKumar last updated on 14/Sep/22

∫(dy/(tan^(−1) y(1+y^2 )))    put         tan^(−1) (y) = x               (dy/(1+y^2 )) = dx  ∫(1/x)dx = ln∣x∣ + C  ∫(dy/(tan^(−1) y(1+y^2 ))) = ln∣tan^(−1) (y)∣ + C

$$\int\frac{\mathrm{dy}}{\mathrm{tan}^{−\mathrm{1}} \mathrm{y}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)} \\ $$$$ \\ $$$${put}\:\:\:\:\:\:\:\:\:{tan}^{−\mathrm{1}} \left({y}\right)\:=\:{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dy}}{\mathrm{1}+{y}^{\mathrm{2}} }\:=\:{dx} \\ $$$$\int\frac{\mathrm{1}}{{x}}{dx}\:=\:{ln}\mid{x}\mid\:+\:\mathrm{C} \\ $$$$\int\frac{\mathrm{dy}}{\mathrm{tan}^{−\mathrm{1}} \mathrm{y}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)}\:=\:{ln}\mid{tan}^{−\mathrm{1}} \left({y}\right)\mid\:+\:\mathrm{C} \\ $$$$ \\ $$

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