Question

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

Answer 1

\[\int\frac{2x - 3}{\left( x^2 - 1 \right) \left( 2x + 3 \right)}dx\]
\[ = \int\frac{\left( 2x - 3 \right)}{\left( x - 1 \right) \left( x + 1 \right) \left( 2x + 3 \right)}dx\]
\[\text{Let }\frac{2x - 3}{\left( x - 1 \right) \left( x + 1 \right) \left( 2x + 3 \right)} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{2x + 3}\]
\[ \Rightarrow \frac{2x - 3}{\left( x - 1 \right) \left( x + 1 \right) \left( 2x + 3 \right)} = \frac{A \left( x + 1 \right) \left( 2x + 3 \right) + B \left( x + 1 \right) \left( 2x + 3 \right) + C \left( x^2 - 1 \right)}{\left( x - 1 \right) \left( x + 1 \right) \left( 2x + 3 \right)}\]
\[ \Rightarrow 2x - 3 = A \left( x + 1 \right) \left( 2x + 3 \right) + B \left( x - 1 \right) \left( 2x + 3 \right) + C \left( x + 1 \right) \left( x - 1 \right) ...........(1)\]
\[\text{Putting }x + 1 = 0\text{ or }x = - 1\text{ in eq. (1)}\]
\[ \Rightarrow - 2 - 3 = B \left( - 1 - 1 \right) \left( - 2 + 3 \right)\]
\[ \Rightarrow - 5 = B \left( - 2 \right) \left( 1 \right)\]
\[ \Rightarrow B = \frac{5}{2}\]
\[\text{Putting }x - 1 = 0\text{ or }x = 1\text{ in eq. (1)}\]
\[ \Rightarrow 2 - 3 = A \left( 1 + 1 \right) \left( 2 + 3 \right)\]
\[ \Rightarrow - 1 = A \left( 2 \right) \left( 5 \right)\]
\[ \Rightarrow A = \frac{- 1}{10}\]
\[\text{Putting }2x + 3 = 0\text{ or }x = \frac{- 3}{2}\text{ in eq. (1)}\]
\[ \Rightarrow 2 \times - \frac{3}{2} - 3 = A \times 0 + B \times 0 + C\left( - \frac{3}{2} + 1 \right) \left( \frac{- 3}{2} - 1 \right)\]
\[ \Rightarrow - 6 = C \left( - \frac{1}{2} \right) \left( \frac{- 5}{2} \right)\]
\[ \Rightarrow C = - \frac{24}{5}\]
\[ \therefore \frac{2x - 3}{\left( x - 1 \right) \left( x + 1 \right) \left( 2x + 3 \right)} = \frac{- 1}{10 \left( x - 1 \right)} + \frac{5}{2 \left( x + 1 \right)} - \frac{24}{5 \left( 2x + 3 \right)}\]
\[ \Rightarrow \int\frac{\left( 2x - 3 \right)}{\left( x - 1 \right) \left( x + 1 \right) \left( 2x + 3 \right)} dx = \frac{- 1}{10}\int\frac{1}{x - 1}dx + \frac{5}{2}\int\frac{1}{x + 1}dx - \frac{24}{5}\int\frac{1}{2x + 3}dx\]
\[ = \frac{- 1}{10} \ln \left| x - 1 \right| + \frac{5}{2} \ln \left| x + 1 \right| - \frac{24}{5} \ln \frac{\left| 2x + 3 \right|}{3} + C\]
\[ = - \frac{1}{10} \ln \left| x - 1 \right| + \frac{5}{2} \ln \left| x + 1 \right| - \frac{12}{5} \ln \left| 2x + 3 \right| + C\]