Question

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

Answer 1

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

\[\text{Let }\frac{1}{x\left( x - 2 \right)\left( x - 4 \right)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x - 4}\]

\[ \Rightarrow \frac{1}{x\left( x - 2 \right)\left( x - 4 \right)} = \frac{A\left( x - 2 \right)\left( x - 4 \right) + B\left( x \right)\left( x - 4 \right) + Cx \cdot \left( x - 2 \right)}{x\left( x - 2 \right)\left( x - 4 \right)}\]

\[ \Rightarrow 1 = A\left( x - 2 \right)\left( x - 4 \right) + B\left( x \right) \cdot \left( x - 4 \right) + Cx . \left( x - 2 \right) ...........(1)\]

\[\text{Putting }x = 0\text{ in eq. (1)}\]

\[ \Rightarrow 1 = A\left( 0 - 2 \right)\left( 0 - 4 \right) + B \times 0 + C \times 0\]

\[ \Rightarrow \frac{1}{8} = A\]

\[\text{Putting }\left( x - 2 \right) = 0\text{ or }x = 2\text{ in eq. (1)}\]

\[ \Rightarrow 1 = A \times 0 + B\left( 2 \right)\left( 2 - 4 \right) + C \times 2 \times 0\]

\[ \Rightarrow B = - \frac{1}{4}\]

\[\text{Putting }\left( x - 4 \right) = 0\text{ or }x = 4\text{ in eq (1)}\]

\[ \Rightarrow 1 = A \times 0 + B \times 0 + C \cdot 4\left( 4 - 2 \right)\]

\[ \Rightarrow C = \frac{1}{8}\]

\[ \therefore \frac{1}{x\left( x - 2 \right)\left( x - 4 \right)} = \frac{1}{8x} - \frac{1}{4\left( x - 2 \right)} + \frac{1}{8\left( x - 4 \right)}\]

\[ \Rightarrow \int\frac{dx}{x\left( x - 2 \right)\left( x - 4 \right)} = \frac{1}{8}\int\frac{1}{x}dx - \frac{1}{4}\int\frac{1}{x - 2}dx + \frac{1}{8}\int\frac{1}{x - 4}dx\]

\[ = \frac{1}{8} \ln \left| x \right| - \frac{1}{4} \ln \left| x - 2 \right| + \frac{1}{8} \ln \left| x - 4 \right| + C\]

\[ = \frac{1}{8}\left( \ln \left| x \right| + \ln \left| x - 4 \right| - 2 \ln \left| x - 2 \right| \right) + C\]

\[ = \frac{1}{8}\left[ \ln \left| \frac{x\left( x - 4 \right)}{\left( x - 2 \right)^2} \right| \right] + C\]