Question

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

Answer 1

We have,

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

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

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

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

Putting\ x - 1 = 0

\[ \Rightarrow x = 1\]

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

\[ \Rightarrow 1 = A \times 6\]

\[ \Rightarrow A = \frac{1}{6}\]

Putting x + 1 = 0

\[ \Rightarrow x = - 1\]

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

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

Putting x + 2 = 0

\[ \Rightarrow x = - 2\]

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

\[ \Rightarrow 1 = C \times 3\]

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

\[ \therefore I = \frac{1}{6}\int\frac{dx}{x - 1} - \frac{1}{2}\int\frac{dx}{x + 1} + \frac{1}{3}\int\frac{dx}{x + 2}\]

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

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

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

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