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Question
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Solution
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`
\[1 = A \left( a^2 - b^2 \right) + B \times 0\]
\[ \Rightarrow A = \frac{1}{a^2 - b^2}\]
\[I = \frac{1}{2}\int\frac{dt}{\left( t - a^2 \right) \left( t - b^2 \right)}\]
\[ = \frac{1}{2 \left( a^2 - b^2 \right)}\int\frac{dt}{t - a^2} + \frac{1}{2 \left( b^2 - a^2 \right)}\int\frac{dt}{t - b^2}\]
\[ = \frac{1}{2 \left( a^2 - b^2 \right)} \log \left| t - a^2 \right| + \frac{1}{2 \left( b^2 - a^2 \right)} \log \left| t - b^2 \right| + C\]
\[ = \frac{1}{2 \left( a^2 - b^2 \right)} \left[ \log \left| t - a^2 \right| - \log \left| t - b^2 \right| \right] + C\]
\[ = \frac{1}{2 \left( a^2 - b^2 \right)} \left[ \log \left| \frac{t - a^2}{t - b^2} \right| \right] + C\]
\[ = \frac{1}{2 \left( a^2 - b^2 \right)} \log \left| \frac{x^2 - a^2}{x^2 - b^2} \right| + C\]
