Question

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

Answer 1

\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)}dx\]
\[\text{Let }\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3}\]
\[ \Rightarrow \frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} = \frac{A \left( x - 2 \right) \left( x - 3 \right) + B \left( x - 1 \right) \left( x - 3 \right) + C \left( x - 1 \right) \left( x - 2 \right)}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)}\]
\[ \Rightarrow x^2 = A \left( x - 2 \right) \left( x - 3 \right) + B \left( x - 1 \right) \left( x - 3 \right) + C \left( x - 1 \right) \left( x - 2 \right) ............(1)\]
\[\text{Putting }x - 1 = 0\text{ or }x = 1\text{ in eq (1)}\]
\[ \Rightarrow 1 = A \left( 1 - 2 \right) \left( 1 - 3 \right)\]
\[ \Rightarrow 1 = A \left( - 1 \right) \left( - 2 \right)\]
\[ \Rightarrow A = \frac{1}{2}\]
\[\text{Putting }x - 2 = 0\text{ or }x = 2\text{ in eq (1)}\]
\[ \Rightarrow 4 = B \left( 2 - 1 \right) \left( 2 - 3 \right)\]
\[ \Rightarrow B = - 4\]
\[\text{Putting }x - 3 = 0\text{ or }x = 3\text{ in eq (1)}\]
\[ \Rightarrow 9 = C \left( 3 - 1 \right) \left( 3 - 2 \right)\]
\[ \Rightarrow C = \frac{9}{2}\]
\[ \therefore \frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} = \frac{1}{2 \left( x - 1 \right)} - \frac{4}{x - 2} + \frac{9}{2 \left( x - 3 \right)}\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)}dx = \frac{1}{2}\int\frac{1}{x - 1}dx - 4\int\frac{1}{x - 2}dx + \frac{9}{2}\int\frac{1}{x - 3}dx\]
\[ = \frac{1}{2}\ln \left| x - 1 \right| - 4 \ln \left| x - 2 \right| + \frac{9}{2} \ln\left| x - 3 \right| + C\]