Question

\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]

Answer 1

\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)}dx\]
\[\text{Let }\frac{5x}{\left( x + 1 \right) \left( x - 2 \right) \left( x + 2 \right)} = \frac{A}{x + 1} + \frac{B}{x - 2} + \frac{C}{x + 2}\]
\[ \Rightarrow \frac{5x}{\left( x + 1 \right) \left( x - 2 \right) \left( x + 2 \right)} = \frac{A \left( x - 2 \right) \left( x + 2 \right) + B \left( x + 1 \right) \left( x + 2 \right) + C \left( x + 1 \right) \left( x - 2 \right)}{\left( x + 1 \right) \left( x - 2 \right) \left( x + 2 \right)}\]
\[ \Rightarrow 5x = A \left( x - 2 \right) \left( x + 2 \right) + B \left( x + 1 \right) \left( x + 2 \right) + C \left( x + 1 \right) \left( x - 2 \right)...........(1)\]
\[\text{Putting }x - 2 = 0\text{ or }x = 2\text{ in eq. (1)}\]
\[ \Rightarrow 5 \times 2 = B \left( 2 + 1 \right) \left( 2 + 2 \right)\]
\[ \Rightarrow B = \frac{10}{3 \times 4}\]
\[ = \frac{5}{6}\]
\[\text{Putting }x + 2 = 0\text{ or }x = - 2\text{ in eq. (1)}\]
\[ \Rightarrow 5 \times - 2 = C \left( - 2 + 1 \right) \left( - 2 - 2 \right)\]
\[ \Rightarrow \frac{- 10}{- 1 \times - 4} = C\]
\[ \Rightarrow C = \frac{- 5}{2}\]
\[\text{Putting }x + 1 = 0\text{ or }x = - 1\text{ in eq. (1)}\]
\[ \Rightarrow - 5 = A \left( - 1 - 2 \right) \left( - 1 + 2 \right)\]
\[ \Rightarrow \frac{- 5}{- 3} = A\]
\[ \Rightarrow A = \frac{5}{3}\]
\[ \therefore \frac{5x}{\left( x + 1 \right) \left( x - 2 \right) \left( x + 2 \right)} = \frac{5}{3} \times \frac{1}{x + 1} + \frac{5}{6 \left( x - 2 \right)} - \frac{5}{2 \left( x + 2 \right)}\]
\[ \Rightarrow \frac{5x}{\left( x + 1 \right) \left( x - 2 \right) \left( x + 2 \right)} = \frac{5}{6} \times \frac{2}{x + 1} + \frac{5}{6 \left( x - 2 \right)} - \frac{5}{6} \left( \frac{3}{x + 2} \right)\]
\[ \therefore \int\frac{5x}{\left( x + 1 \right) \left( x - 2 \right) \left( x + 2 \right)}dx = \frac{5}{6}\int\frac{2}{x + 1} dx + \frac{5}{6}\int\frac{1}{x - 2}dx - \frac{5}{6}\int\frac{3}{x + 2} dx\]
\[ = \frac{5}{6}\left[ 2 \ln \left| x + 1 \right| + \ln \left| x - 2 \right| - 3 \ln \left| x + 2 \right| \right] + C\]
\[ = \frac{5}{6} \left[ \ln \left| x + 1 \right|^2 + \ln \left| x - 2 \right| - \ln \left| x + 2 \right|^3 \right] + C\]
\[ = \frac{5}{6} \ln \left| \frac{\left( x + 1 \right)^2 \left( x - 2 \right)}{\left( x + 2 \right)^3} \right| + C\]