Question

Evaluate the following integral:

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

Answer 1

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

We express

\[\frac{3x - 2}{\left( x + 1 \right)^2 \left( x + 3 \right)} = \frac{A}{x + 1} + \frac{B}{\left( x + 1 \right)^2} + \frac{C}{x + 3}\]

\[ \Rightarrow 3x - 2 = A\left( x + 1 \right)\left( x + 3 \right) + B\left( x + 3 \right) + C \left( x + 1 \right)^2 \]

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

\[0 = A + C\text{ and }3 = 4A + B + 2C\text{ and }- 2 = 3A + 3B + C\]

\[\text{or }A = \frac{11}{4}\text{ and }B = - \frac{5}{2}\text{ and }C = - \frac{11}{4}\]

\[ \therefore I = \int\left( \frac{\frac{11}{4}}{x + 1} + \frac{- \frac{5}{2}}{\left( x + 1 \right)^2} + \frac{- \frac{11}{4}}{x + 3} \right)dx\]

\[ = \frac{11}{4}\int\frac{1}{x + 1}dx - \frac{5}{2}\int\frac{1}{\left( x + 1 \right)^2} dx - \frac{11}{4}\int\frac{1}{x + 3} dx\]

\[ = \frac{11}{4}\log\left| x + 1 \right| + \frac{5}{2\left( x + 1 \right)} - \frac{11}{4}\log\left| x + 3 \right| + c\]

\[\text{Hence, }\int\frac{3x - 2}{\left( x + 1 \right)^2 \left( x + 3 \right)}dx = \frac{11}{4}\log\left| x + 1 \right| + \frac{5}{2\left( x + 1 \right)} - \frac{11}{4}\log\left| x + 3 \right| + c\]