Question

Evaluate the following integral:

\[\int\frac{x^2 + 1}{\left( x^2 + 4 \right)\left( x^2 + 25 \right)}dx\]

Answer 1

\[\text{Let }I = \int\frac{x^2 + 1}{\left( x^2 + 4 \right)\left( x^2 + 25 \right)}dx\]

We express

\[\frac{x^2 + 1}{\left( x^2 + 4 \right)\left( x^2 + 25 \right)} = \frac{Ax + B}{x^2 + 4} + \frac{Cx + D}{x^2 + 25}\]

\[ \Rightarrow x^2 + 1 = \left( Ax + B \right)\left( x^2 + 25 \right) + \left( Cx + D \right)\left( x^2 + 4 \right)\]

Equating the coefficients of `x^3 , x^2 , x` and constants, we get

\[0 = A + C\text{ and }1 = B + D\text{ and }0 = 25A + 4C\text{ and }1 = 25B + 4D\]

\[\text{or }A = 0\text{ and }B = - \frac{1}{7}\text{ and }C = 0\text{ and }D = \frac{8}{7}\]

\[ \therefore I = \int\left( \frac{- \frac{1}{7}}{x^2 + 4} + \frac{\frac{8}{7}}{x^2 + 25} \right)dx\]

\[ = - \frac{1}{7}\int\frac{1}{x^2 + 4}dx + \frac{8}{7}\int\frac{1}{x^2 + 25} dx\]

\[ = - \frac{1}{7} \times \frac{1}{2} \tan^{- 1} \frac{x}{2} + \frac{8}{7} \times \frac{1}{5} \tan^{- 1} \frac{x}{5} + c\]

\[ = - \frac{1}{14} \tan^{- 1} \frac{x}{2} + \frac{8}{35} \tan^{- 1} \frac{x}{5} + c\]

\[\text{Hence, }\int\frac{x^2 + 1}{\left( x^2 + 4 \right)\left( x^2 + 25 \right)}dx = - \frac{1}{14} \tan^{- 1} \frac{x}{2} + \frac{8}{35} \tan^{- 1} \frac{x}{5} + c\]