Question

Evaluate the following integrals:

\[\int\frac{1}{\left( x^2 + 2x + 10 \right)^2}dx\]

 

Answer 1

\[\text{Let I }= \int\frac{1}{\left( x^2 + 2x + 10 \right)^2}dx\]

\[ = \int\frac{1}{\left[ \left( x + 1 \right)^2 + 3^2 \right]^2}dx\]

\[ \text{Let x + 1 }= 3\tan\theta\]

\[ \text{On differentiating both sides, we get}\]

\[ dx = 3 \sec^2 \theta \text{ dθ }\]

\[ \therefore I = \int\frac{1}{\left[ 3^2 \tan^2 \theta + 3^2 \right]^2}3 \sec^2  θ    \text{ dθ }\]

\[ = \frac{1}{27}\int\frac{\sec^2 \theta}{\sec^4 \theta}d\theta\]

\[ = \frac{1}{27}\int\frac{1}{\sec^2 \theta}d\theta\]

`= {1}/{27}\int \text{ cos}^2  θ  \text{ dθ }`

 

` = {1}/{54}\int\left( 1 + cos2θ  ) dθ `

\[ = \frac{1}{54}\left( \theta + \frac{\sin2\theta}{2} \right) + c\]

\[ = \frac{1}{54}\left( \theta + \frac{\tan\theta}{1 + \tan^2 \theta} \right) + c\]

\[ = \frac{1}{54}\left( \tan^{- 1} \frac{x + 1}{3} + \frac{\tan\left( \tan^{- 1} \frac{x + 1}{3} \right)}{1 + \tan^2 \left( \tan^{- 1} \frac{x + 1}{3} \right)} \right) + c\]

\[ = \frac{1}{54}\left( \tan^{- 1} \frac{x + 1}{3} + \frac{\frac{x + 1}{3}}{1 + \left( \frac{x + 1}{3} \right)^2} \right) + c\]

\[ = \frac{1}{54}\left( \tan^{- 1} \frac{x + 1}{3} + \frac{\frac{x + 1}{3}}{\frac{x^2 + 2x + 10}{9}} \right) + c\]

\[ = \frac{1}{54}\left( \tan^{- 1} \frac{x + 1}{3} + \frac{3\left( x + 1 \right)}{x^2 + 2x + 10} \right) + c\]

\[Hence, \int\frac{1}{\left( x^2 + 2x + 10 \right)^2}dx = \frac{1}{54}\left( \tan^{- 1} \frac{x + 1}{3} + \frac{3\left( x + 1 \right)}{x^2 + 2x + 10} \right) + c\]