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Question Number 190740 by mathlove last updated on 10/Apr/23

$$\int \frac{dx}{{(x^2+5)}^2}=?$$

Answered by gatocomcirrose last updated on 10/Apr/23

$$\mathrm{x}=\sqrt{5}\mathrm{tg}\theta \Rightarrow \mathrm{dx}=\sqrt{5}{\sec }^2\theta \mathrm{d}\theta$$$$\int \frac{\sqrt{5}{\sec }^2\theta }{{25\sec }^4\theta }\mathrm{d}\theta =\frac{\sqrt{5}}{25}\int {\cos }^2\theta \mathrm{d}\theta$$$$=\frac{\sqrt{5}}{50}\int \cos \left(2\theta \right)+1\mathrm{d}\theta =\frac{\sqrt{5}}{100}\sin \left(2\theta \right)+\theta$$$$=\frac{\mathrm{x}}{10({\mathrm{x}}^2+5)}+\arctan \left(\frac{\mathrm{x}}{\sqrt{5}}\right)+\mathrm{C}$$

Commented by mehdee42 last updated on 10/Apr/23

$$verygood$$$$decreasingorder$$$$if{I}_n=\int \frac{dx}{{(x^2+a)}^n}\Rightarrow I_{n+1}=\frac{x}{2na{(x^2+a)}^n}+\frac{2n-1}{2na}{I}_n$$

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