Question

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

Answer 1

We have,

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

\[\text{Let }\frac{2x + 1}{\left( x + 2 \right) \left( x - 3 \right)^2} = \frac{A}{x + 2} + \frac{B}{x - 3} + \frac{C}{\left( x - 3 \right)^2}\]

\[ \Rightarrow \frac{2x + 1}{\left( x + 2 \right) \left( x - 3 \right)^2} = \frac{A \left( x - 3 \right)^2 + B \left( x + 2 \right) \left( x - 3 \right) + C \left( x + 2 \right)}{\left( x + 2 \right) \left( x - 3 \right)^2}\]

\[ \Rightarrow 2x + 1 = A \left( x^2 - 6x + 9 \right) + B \left( x^2 - x - 6 \right) + C \left( x + 2 \right)\]

\[ \Rightarrow 2x + 1 = \left( A + B \right) x^2 + \left( - 6A - B + C \right) x + \left( 9A - 6B + 2C \right)\]

Equating the coefficients of like terms

\[A + B = 0 ..................(1)\]

\[ - 6A - B + C = 2 ....................(2)\]

\[9A - 6B + 2C = 1 .......................(3)\]

Solving (1), (2) and (3), we get

\[A = - \frac{3}{25}, B = \frac{3}{25}\text{ and }C = \frac{7}{5}\]

\[ \therefore \frac{\left( 2x + 1 \right) dx}{\left( x + 2 \right) \left( x - 3 \right)^2} = - \frac{3}{25 \left( x + 2 \right)} + \frac{3}{25 \left( x - 3 \right)} + \frac{7}{5 \left( x - 3 \right)^2}\]

\[ \Rightarrow I = - \frac{3}{25}\int\frac{dx}{x + 2} + \frac{3}{25}\int\frac{dx}{x - 3} + \frac{7}{5}\int \left( x - 3 \right)^{- 2} dx\]

\[ = - \frac{3}{25} \log \left| x + 2 \right| + \frac{3}{25} \log \left| x - 3 \right| + \frac{7}{5}\left[ \frac{\left( x - 3 \right)^{- 1}}{- 1} \right] + C\]

\[ = - \frac{3}{25}\log \left| x + 2 \right| + \frac{3}{25} \log \left| x - 3 \right| - \frac{7}{5 \left( x - 3 \right)} + C\]