Question

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

Answer 1

We have,

\[I = \int\frac{x dx}{\left( x^2 - a^2 \right) \left( x^2 - b^2 \right)}\]

Putting `x^2 = t`

\[ \Rightarrow 2x\ dx = dt\]

\[ \Rightarrow x\ dx = \frac{dt}{2}\]

\[ \therefore I = \frac{1}{2}\int\frac{dt}{\left( t - a^2 \right) \left( t - b^2 \right)}\]

\[\text{Let }\frac{1}{\left( t - a^2 \right) \left( t - b^2 \right)} = \frac{A}{t - a^2} + \frac{B}{t - b^2}\]

\[ \Rightarrow \frac{1}{\left( t - a^2 \right) \left( t - b^2 \right)} = \frac{A \left( t - b^2 \right) + B \left( t - a^2 \right)}{\left( t - a^2 \right) \left( t - b^2 \right)}\]

\[ \Rightarrow 1 = A \left( t - b^2 \right) + B \left( t - a^2 \right)\]

Putting `t = b^2`

\[1 = A \times 0 + B \left( b^2 - a^2 \right)\]

\[ \Rightarrow B = \frac{1}{b^2 - a^2}\]

Putting `t = a^2`

\[1 = A \left( a^2 - b^2 \right) + B \times 0\]

\[ \Rightarrow A = \frac{1}{a^2 - b^2}\]

\[I = \frac{1}{2}\int\frac{dt}{\left( t - a^2 \right) \left( t - b^2 \right)}\]

\[ = \frac{1}{2 \left( a^2 - b^2 \right)}\int\frac{dt}{t - a^2} + \frac{1}{2 \left( b^2 - a^2 \right)}\int\frac{dt}{t - b^2}\]

\[ = \frac{1}{2 \left( a^2 - b^2 \right)} \log \left| t - a^2 \right| + \frac{1}{2 \left( b^2 - a^2 \right)} \log \left| t - b^2 \right| + C\]

\[ = \frac{1}{2 \left( a^2 - b^2 \right)} \left[ \log \left| t - a^2 \right| - \log \left| t - b^2 \right| \right] + C\]

\[ = \frac{1}{2 \left( a^2 - b^2 \right)} \left[ \log \left| \frac{t - a^2}{t - b^2} \right| \right] + C\]

\[ = \frac{1}{2 \left( a^2 - b^2 \right)} \log \left| \frac{x^2 - a^2}{x^2 - b^2} \right| + C\]