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Question Number 101601 by Dwaipayan Shikari last updated on 03/Jul/20

$${\int }_{\sqrt{2}-1}^{\sqrt{2}+1}\frac{x^4+x^2+1}{{(x^2+1)}^2}dx$$

Answered by bemath last updated on 03/Jul/20

$$\int \frac{{({\mathrm{x}}^2+1)}^2-{\mathrm{x}}^2}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}=\mathrm{x}-\int \frac{{\mathrm{x}}^2}{{({\mathrm{x}}^2+1)}^2}\mathrm{dx}$$$${\mathrm{I}}_2=\int \frac{x^2}{{(x^2+1)}^2}dx$$$$[x=\tan p]$$$${\mathrm{I}}_2=\int \frac{\tan {}^{2}p.\sec {}^{2}pdp}{\sec {}^{4}p}$$$$=\int \tan {}^{2}p\cos {}^{2}pdp$$$$=\int (\frac{1}{2}-\frac{1}{2}\cos 2p)dp$$$$=\frac{1}{2}p-\frac{1}{4}\sin 2p=\frac{1}{2}{\tan }^{-1}\left(x\right)-\frac{x}{2(x^2+1)}$$$$I=2-\frac{1}{2}({\tan }^{-1}(\sqrt{2}+1)-{\tan }^{-1}(\sqrt{2}-1))$$$$-\frac{1}{2}(\frac{\sqrt{2}+1}{4+2\sqrt{2}}-\frac{\sqrt{2}-1}{4-2\sqrt{2}})$$

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