Evaluate the following integrals: 1 (x^2+2x+10)^2 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{(\text{x}^2+2\text{x}+10)^2}\text{ dx}$

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Answer

Let $\int\frac{1}{(\text{x}^2+2\text{x}+10)^2}\text{ dx}$
$=\int\frac{1}{\big[(\text{x}+1)^2+3^2\big]}\text{ dx}$
Let $\text{x}+1=3\tan\theta$
On differentiating both sides, we get
$\text{dx}=3\sec^2\theta\text{ d}\theta$
$\therefore\ \text{I}=\int\frac{1}{\big[3^2\tan^2\theta+3^2\big]^2}3\sec^2\theta\text{ d}\theta$
$=\frac{1}{27}\int\frac{\sec^2\theta}{\sec^4\theta}\text{ d}\theta$
$=\frac{1}{27}\int\frac{1}{\sec^2\theta}\text{ d}\theta$
$=\frac{1}{27}\int\cos^2\theta\text{ d}\theta$
$=\frac{1}{54}\int(1+\cos2\theta)\text{d}\theta$
$=\frac{1}{54}\Big(\theta+\frac{\sin2\theta}{2}\Big)+\text{ C}$
$=\frac{1}{54}\Big(\theta+\frac{\tan\theta}{1+\tan^2\theta}\Big)+\text{C}$
$=\frac{1}{54}\begin{pmatrix}\tan^{-1}\frac{\text{x}+1}{3}+\frac{\tan\Big(\tan^{-1}\frac{\text{x}+1}{3}\Big)}{1+\tan^{2}\Big(\tan^{-1}\frac{\text{x}+1}{3}\Big)}\end{pmatrix}+\text{C}$
$=\frac{1}{54}\begin{pmatrix}\tan^{-1}\frac{\text{x}+1}{3}+\frac{\frac{\text{x}+1}{3}}{1+\Big(\frac{\text{x}+1}{3}\Big)^2}\end{pmatrix}+\text{C}$
$=\frac{1}{54}\Bigg(\tan^{-1}\frac{\text{x}+1}{3}+\frac{\frac{\text{x}+1}{3}}{\frac{\text{x}^2+2\text{x}+10}{9}}\Bigg)+\text{C}$
$=\frac{1}{54}\bigg(\tan^{-1}\frac{\text{x}+1}{3}+\frac{3(\text{x}+1)}{\text{x}^2+2\text{x}+10}\bigg)+\text{C}$

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