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

∫(u^2 /((1+u^2 )^2 ))du

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

Answered by liberty last updated on 23/Jan/21

J=∫ (((u^2 +1)−1)/((1+u^2 )^2 )) du = ∫ (du/(1+u^2 ))−∫(du/((1+u^2 )^2 )) J_1 =∫ (du/(1+u^2 )) = tan^(−1) (u)+c_1 J_2 =∫ (du/((1+u^2 )^2 )) ; u = tan q J_2 =∫ ((sec^2 q dq)/(sec^4 q)) = ∫ cos^2 q dq J_2 = (1/2)q+(1/4)sin 2q+c_2 J_2 =((tan^(−1) (u))/2)+(u/(2(1+u^2 )))+c_2 ∴ J= (1/2)tan^(−1) (u)−(1/2) (u/((1+u)^2 ))+C

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

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