Question

$\int \frac{x}{(x^2-a^2)(x^2-b^2)}dx$ is equal to $\frac{1}{k(b^2-a^2)}\ln \frac{(x^2-b^2)}{(x^2-a^2)}$. Find $k$

Answer 1

Now,

$\int \frac{x}{(x^2-a^2)(x^2-b^2)}dx$

$=\frac{1}{2(b^2-a^2)}\int [\frac{1}{(x^2-b^2)}-\frac{1}{(x^2-a^2)}]\left(2x\right)dx$

$=\frac{1}{2(b^2-a^2)}\int [\frac{d(x^2-b^2)}{(x^2-b^2)}-\frac{d(x^2-a^2)}{(x^2-a^2)}]$

$=\frac{1}{2(b^2-a^2)}[\log (x^2-b^2)-\log (x^2-a^2)]+c$ [ $c$ is integrating constant]

$=\frac{1}{2(b^2-a^2)}\log \frac{x^2-b^2}{x^2-a^2}+c$

Comparing this to the given problem, we get, $k=2$.