Question

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

Answer 1

\[\int\frac{\left( 2x + 1 \right)}{\left( x + 1 \right)\left( x - 2 \right)} dx \]
\[\text{Let }\frac{2x + 1}{\left( x + 1 \right)\left( x - 2 \right)} = \frac{A}{x + 1} + \frac{B}{x - 2} .........(1)\]
\[ \Rightarrow \frac{2x + 1}{\left( x + 1 \right)\left( x - 2 \right)} = \frac{A\left( x - 2 \right) + B\left( x + 1 \right)}{\left( x + 1 \right)\left( x - 2 \right)}\]
\[\text{Then, }\left( 2x + 1 \right) = A\left( x - 2 \right) + B\left( x + 1 \right) ............(2)\]
\[\text{Putting }\left( x - 2 \right) = 0\text{ or }x = 2\text{ in eq. (2) }\]
\[ \Rightarrow 2 \times 2 + 1 = A \times 0 + B\left( 2 + 1 \right)\]
\[ \Rightarrow B = \frac{5}{3}\]
\[\text{Putting }\left( x + 1 \right) = 0\text{ or }x = - 1\text{ in eq. (2)} \]
\[2 \times - 1 + 1 + A\left( - 1 - 2 \right) + B \times 0\]
\[ \Rightarrow - 1 = A\left( - 3 \right)\]
\[ \Rightarrow A = \frac{1}{3}\]
\[\text{Substituting the values of A and B in eq. (1) , we get} \]
\[ \therefore \frac{2x + 1}{\left( x + 1 \right)\left( x - 2 \right)} = \frac{1}{3}\left( x + 1 \right) + \frac{5}{3}\left( x - 2 \right)\]
\[\int\frac{\left( 2x + 1 \right)dx}{\left( x + 1 \right)\left( x - 2 \right)} = \frac{1}{3}\int\frac{1}{x + 1}dx + \frac{5}{3}\int\frac{1}{x - 2}dx\]
\[ = \frac{1}{3} \ln \left| x + 1 \right| + \frac{5}{3} \ln \left| x - 2 \right| + C\]