Question

\[\int\frac{3 + 4x - x^2}{\left( x + 2 \right) \left( x - 1 \right)} dx\]

Answer 1

\[\int\left( \frac{3 + 4x - x^2}{x^2 - x + 2x - 2} \right)dx\]
\[ = \int\left( \frac{- x^2 + 4x + 3}{x^2 + x - 2} \right)dx\]
\[\frac{- x^2 + 4x + 3}{x^2 + x - 2} = - 1 + \frac{5x + 1}{x^2 + x - 2} .............(1)\]
\[ \therefore \frac{5x + 1}{x^2 + x - 2} = \frac{5x + 1}{x^2 + 2x - x - 2}\]
\[ = \frac{5x + 1}{x\left( x + 2 \right) - 1\left( x + 2 \right)}\]
\[\text{Let }\frac{5x + 1}{\left( x - 1 \right)\left( x + 2 \right)} = \frac{A}{x - 1} + \frac{B}{x + 2}\]
\[ \Rightarrow \frac{5x + 1}{\left( x - 1 \right)\left( x + 2 \right)} = \frac{A\left( x + 2 \right) + B\left( x - 1 \right)}{\left( x - 1 \right)\left( x + 2 \right)}\]
\[ \Rightarrow 5x + 1 = A\left( x + 2 \right) + B\left( x - 1 \right) ..........(2)\]
\[\text{Putting }x + 2 = 0\text{ or }x = - 2\text{ in eq. (2)}\]
\[ \Rightarrow 5x - 2 + 1 = A \times 0 + B\left( - 2 - 1 \right)\]
\[ \Rightarrow B = 3\]
\[\text{Putting }x - 1 = 0\text{ or }x = 1\text{ in eq. (2)}\]
\[ \Rightarrow 5 \times 1 + 1 = A\left( 3 \right) + B \times 0\]
\[ \Rightarrow A = 2\]
\[ \therefore \frac{5x + 1}{\left( x - 1 \right)\left( x + 2 \right)} = \frac{2}{x - 1} + \frac{3}{x + 2} ............(3)\]
From (1) and (3)
\[\frac{- x^2 + 4x + 3}{x^2 + x - 2} = - 1 + \frac{2}{x - 1} + \frac{3}{x + 2}\]
\[ \Rightarrow \int\frac{- x^2 + 4x + 3}{x^2 + x - 2}dx = \int - 1 dx + \int\frac{2}{x - 1}dx + \int\frac{3}{x + 2} dx\]
\[ = - x + 2 \ln\left( x - 1 \right) + 3 \ln\left( x + 2 \right) + C\]