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Question Number 166182 by SANOGO last updated on 14/Feb/22

$$calculerlaprimitivede$$$$\int \frac{t^2}{{(1+t^2)}^2}dt$$

Answered by greogoury55 last updated on 14/Feb/22

$$t=\tan u$$$$Y=\int \frac{\tan {}^{2}u}{{(1+\tan {}^{2}u)}^2}\sec {}^{2}udu$$$$Y=\int \frac{\tan {}^{2}u}{\sec {}^{4}u}\sec {}^{2}udu$$$$Y=\int \frac{\tan {}^{2}u}{\sec {}^{2}u}du=\int \sin {}^{2}udu$$$$Y=\int \frac{1-\cos 2u}{2}du=\frac{u-\frac{1}{2}\sin 2u}{2}+c$$$$Y=\frac{1}{2}\arctan \left(t\right)-\frac{t}{2t^2+2}+c$$$$$$

Commented by SANOGO last updated on 14/Feb/22

$$merci$$

Answered by MJS_new last updated on 15/Feb/22

$$\int \frac{t^2}{{(t^2+1)}^2}dt=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=-\frac{t}{2(t^2+1)}+\frac{1}{2}\int \frac{dt}{t^2+1}=$$$$=-\frac{t}{2(t^2+1)}+\frac{1}{2}\arctan t+C$$

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