Question

Evaluate the following integrals:

$\int \frac{1}{{\left(x^2+2x+10\right)}^2}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{{\left(x^2+2x+10\right)}^2}\mathrm{d}x \\ =\int \frac{1}{{\left[{\left(x+1\right)}^2+3^2\right]}^2}\mathrm{d}x \\ \mathrm{Let}x+1=3\tan \theta \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \mathrm{d}x=3{\sec }^2\theta \mathrm{d}\theta \\ \therefore I=\int \frac{1}{{\left[3^2{\tan }^2\theta +3^2\right]}^2}3{\sec }^2\theta \mathrm{d}\theta \\ =\frac{1}{27}\int \frac{{\sec }^2\theta }{{\sec }^4\theta }\mathrm{d}\theta \\ =\frac{1}{27}\int \frac{1}{{\sec }^2\theta }\mathrm{d}\theta \\ =\frac{1}{27}\int {\cos }^2\theta \mathrm{d}\theta \\ =\frac{1}{54}\int \left(1+\cos 2\theta \right)\mathrm{d}\theta \\ =\frac{1}{54}\left(\theta +\frac{\sin 2\theta }{2}\right)+c \\ =\frac{1}{54}\left(\theta +\frac{\tan \theta }{1+{\tan }^2\theta }\right)+c \\ =\frac{1}{54}\left({\tan }^{-1}\frac{x+1}{3}+\frac{\tan \left({\tan }^{-1}{\displaystyle \frac{x+1}{3}}\right)}{1+{\tan }^2\left({\tan }^{-1}{\displaystyle \frac{x+1}{3}}\right)}\right)+c \\ =\frac{1}{54}\left({\tan }^{-1}\frac{x+1}{3}+\frac{{\displaystyle \frac{x+1}{3}}}{1+{\left({\displaystyle \frac{x+1}{3}}\right)}^2}\right)+c \\ =\frac{1}{54}\left({\tan }^{-1}\frac{x+1}{3}+\frac{{\displaystyle \frac{x+1}{3}}}{{\displaystyle \frac{x^2+2x+10}{9}}}\right)+c \\ =\frac{1}{54}\left({\tan }^{-1}\frac{x+1}{3}+\frac{{\displaystyle 3\left(x+1\right)}}{{\displaystyle x^2+2x+10}}\right)+c \\ \mathrm{Hence},\int \frac{1}{{\left(x^2+2x+10\right)}^2}\mathrm{d}x=\frac{1}{54}\left({\tan }^{-1}\frac{x+1}{3}+\frac{{\displaystyle 3\left(x+1\right)}}{{\displaystyle x^2+2x+10}}\right)+c \end{gathered}$$