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Question Number 93484 by mashallah last updated on 13/May/20

$$\int {\mathrm{t}}^2/{(1+{\mathrm{t}}^2)}^2\mathrm{dx}=$$

Commented by prakash jain last updated on 13/May/20

$$\mathrm{function}\mathrm{is}\mathrm{in}t\mathrm{so}\mathrm{constant}\mathrm{wrt}x$$$$\frac{t^2}{{(1+t^2)}^2}x+C$$

Commented by mathmax by abdo last updated on 13/May/20

$$I=\int \frac{t^2}{{(1+t^2)}^2}dt\Rightarrow I=\int \frac{t^2+1-1}{(1+t^2)}dt=\int \frac{dt}{1+t^2}-\int \frac{dt}{{(1+t^2)}^2}$$$$wehave\int \frac{dt}{1+t^2}=arctan\left(t\right)+c_1$$$$\int \frac{dt}{{(1+t^2)}^2}{=}_{t=tanx}\int \frac{(1+{tan}^{2}x)dx}{{(1+{tan}^{2}x)}^2}=\int {cos}^{2}xdx$$$$=\frac{1}{2}\int (1+cos\left(2x\right))dx=\frac{x}{2}+\frac{1}{4}sin\left(2x\right)+c_2$$$$sin\left(2x\right)=\frac{2tanx}{1+{tan}^{2}x}=\frac{2t}{1+t^2}\Rightarrow \int \frac{dt}{{(1+t^2)}^2}=\frac{1}{2}arctan\left(t\right)+\frac{t}{2(1+t^2)}+c_2$$$$\Rightarrow I=\frac{1}{2}arctan\left(t\right)-\frac{t}{2(1+t^2)}+C$$

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