2-1-2-1-x-4-x-2-1-x-2-1-2-dx-

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}} d x

Answered by bemath last updated on 03/Jul/20

\int \frac{{(x^{2} + 1)}^{2} - x^{2}}{{(x^{2} + 1)}^{2}} dx = x - \int \frac{x^{2}}{{(x^{2} + 1)}^{2}} dx

I_{2} = \int \frac{x^{2}}{{(x^{2} + 1)}^{2}} d x

[x = \tan p]

I_{2} = \int \frac{\tan^{2} p . \sec^{2} p d p}{\sec^{4} p}

= \int \tan^{2} p \cos^{2} p d p

= \int (\frac{1}{2} - \frac{1}{2} \cos 2 p) d p

= \frac{1}{2} p - \frac{1}{4} \sin 2 p = \frac{1}{2} \tan^{- 1} (x) - \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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