Question

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

Answer 1

We have,

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

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

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

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

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

Putting `x - 1 = 0`

\[ \Rightarrow x = 1\]

\[1 + 1 = A \times 0 + B \left( 2 + 1 \right) \left( 1 + 1 \right) + C \times 0\]

\[ \Rightarrow 2 = B\left( 3 \right)\left( 2 \right)\]

\[ \Rightarrow B = \frac{1}{3}\]

Putting `x + 1 = 0`

\[ \Rightarrow x = - 1\]

\[1 + 1 = A \times 0 + B \times 0 + C \left( - 2 + 1 \right) \left( - 1 - 1 \right)\]

\[ \Rightarrow 2 = C \left( - 1 \right) \left( - 2 \right)\]

\[ \Rightarrow C = 1\]

Putting `2x + 1 = 0`

\[ \Rightarrow x = - \frac{1}{2}\]

\[ \left( - \frac{1}{2} \right)^2 + 1 = A \left( \frac{1}{4} - 1 \right)\]

\[ \Rightarrow \frac{1}{4} + 1 = A \left( - \frac{3}{4} \right)\]

\[ \Rightarrow \frac{5}{4} = A \left( - \frac{3}{4} \right)\]

\[A = - \frac{5}{3}\]

\[ \therefore I = - \frac{5}{3}\int\frac{dx}{2x + 1} + \frac{1}{3}\int\frac{dx}{x - 1} + \int\frac{dx}{x + 1}\]

\[ = - \frac{5}{3} \times \frac{\log \left| 2x + 1 \right|}{2} + \frac{1}{3} \log \left| x - 1 \right| + \log \left| x + 1 \right| + C\]

\[ = - \frac{5}{6} \log \left| 2x + 1 \right| + \frac{1}{3} \log \left| x - 1 \right| + \log \left| x + 1 \right| + C\]