Question

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

Answer 1

\[\int\frac{x^3}{\left( x - 1 \right) \left( x - 2 \right)\left( x - 3 \right)}dx\]
\[ = \int\frac{x^3}{\left( x - 1 \right) \left( x^2 - 5x + 6 \right)}dx\]
\[ = \int\frac{x^3}{x^3 - 5 x^2 + 6x - x^2 + 5x - 6}dx\]
\[ = \int\frac{x^3}{x^3 - 6 x^2 + 11x - 6}dx\]
\[ \therefore \frac{x^3}{x^3 - 6 x^2 + 11x - 6} = 1 + \frac{6 x^2 + 11x + 6}{x^2 - 6 x^2 + 11x - 6}\]
\[ \Rightarrow \frac{x^3}{x^3 - 6 x^2 + 11x - 6} = 1 + \frac{6 x^2 - 11x + 6}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)}\]
\[ \therefore \int\frac{x^3}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx = \int dx + \int\frac{\left( 6 x^2 - 11x + 6 \right)}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)}dx ............(1)\]

\[\text{Let }\frac{6 x^2 - 11x + 6}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} = \frac{A}{x - 1} + \frac{B}{x - 2} + \frac{C}{x - 3}\]
\[ \Rightarrow \frac{6 x^2 - 11x + 6}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} = \frac{A \left( x - 2 \right) \left( x - 3 \right) + B \left( x - 1 \right) \left( x - 3 \right) + C \left( x - 1 \right) \left( x - 2 \right)}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)}\]
\[ \Rightarrow 6 x^2 - 11x + 6 = A \left( x - 2 \right) \left( x - 3 \right) + B \left( x - 1 \right) \left( x - 3 \right) + C \left( x - 1 \right) \left( x - 2 \right) ..............(2)\]
\[\text{Putting }x - 2 = 0\text{ or }x = 2\text{ in eq. (2)}\]
\[ \Rightarrow 6 \times 4 - 22 + 6 = B \left( 2 - 1 \right) \left( 2 - 3 \right)\]
\[ \Rightarrow 8 = B \left( - 1 \right)\]
\[ \Rightarrow B = - 8\]
\[\text{Putting }x - 3 = 0\text{ or }x = 3\text{ in eq. (2)}\]
\[ \Rightarrow 6 \times 3^2 - 11 \times 3 + 6 = C \left( 3 - 1 \right) \left( 3 - 2 \right)\]
\[ \Rightarrow 27 = C \left( 2 \right) \left( 1 \right)\]
\[ \Rightarrow C = \frac{27}{2}\]
\[\text{Putting }x - 1 = 0\text{ or }x = 1\text{ in eq. (2)}\]
\[ \Rightarrow 6 \times 1 - 11 + 6 = A \left( 1 - 2 \right) \left( 1 - 3 \right)\]
\[ \Rightarrow 1 = A \left( - 1 \right) \left( - 2 \right)\]
\[ \Rightarrow A = \frac{1}{2}\]
\[ \therefore \frac{6 x^2 - 11x + 6}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} = \frac{1}{2\left( x - 1 \right)} - \frac{8}{x - 2} + \frac{27}{2\left( x - 3 \right)}..........(3)\]
From eq. (2) and (3)
\[ \therefore \int\frac{x^3 dx}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} = \int dx + \frac{1}{2}\int\frac{1}{x - 1}dx - 8\int\frac{1}{x - 2}dx + \frac{27}{2}\int\frac{1}{x - 3}dx\]
\[ = x + \frac{1}{2} \ln \left| x - 1 \right| - 8 \ln \left| x - 2 \right| + \frac{27}{2} \ln \left| x - 3 \right| + C\]