Question

\[\int\frac{x^2 + x - 1}{x^2 + x - 6} dx\]

Answer 1

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

\[\Rightarrow 1 = A \times 0 + B\left( - 3 - 2 \right)\]

\[\Rightarrow B = - \frac{1}{5}\]

\[\text{Putting }x - 2 = 0\text{ or }x = 2\text{ in eq. (2)}\]

\[\Rightarrow 1 = A\left( 2 + 3 \right) + B \times 0\]

\[\Rightarrow A = \frac{1}{5}\]

\[\therefore \frac{1}{\left( x - 2 \right)\left( x + 3 \right)} = \frac{1}{5}\left( x - 2 \right) - \frac{1}{5}\left( x + 3 \right)\]

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

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

\[ = \frac{1}{5} \ln \left| \frac{x - 2}{x + 3} \right| + C ...........(3)\]

From eq. (1) and eq. (3)

\[ \therefore \int\left( \frac{x^2 + x - 1}{x^2 + x - 6} \right)dx = x + \frac{5}{5} \ln \left| \frac{x - 2}{x + 3} \right| + C\]

\[ = x + \ln \left| x - 2 \right| - \ln \left| x + 3 \right| + C\]