int frac{x , dx}{(x^2 - a^2)(x^2 - b^2)} = | vedclass

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(C) Let $I = \int \frac{x \, dx}{(x^2 - a^2)(x^2 - b^2)}$.Put $x^2 = t$,then $2x \, dx = dt$,or $x \, dx = \frac{dt}{2}$.Substituting this into the in

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

Vedclass · Answer

(C) Let $I = \int \frac{x \, dx}{(x^2 - a^2)(x^2 - b^2)}$.Put $x^2 = t$,then $2x \, dx = dt$,or $x \, dx = \frac{dt}{2}$.Substituting this into the integral,we get:$I = \int \frac{dt}{2(t - a^2)(t - b^2)} = \frac{1}{2} \int \frac{dt}{(t - a^2)(t - b^2)}$.Using partial fractions:$\frac{1}{(t - a^2)(t - b^2)} = \frac{1}{a^2 - b^2} \left( \frac{1}{t - a^2} - \frac{1}{t - b^2} \right)$.Therefore,$I = \frac{1}{2(a^2 - b^2)} \int \left( \frac{1}{t - a^2} - \frac{1}{t - b^2} \right) dt$.$I = \frac{1}{2(a^2 - b^2)} [\log |t - a^2| - \log |t - b^2|] + c$.$I = \frac{1}{2(a^2 - b^2)} \log \left| \frac{t - a^2}{t - b^2} \right| + c$.Substituting $t = x^2$ back,we get:$I = \frac{1}{2(a^2 - b^2)} \log \left| \frac{x^2 - a^2}{x^2 - b^2} \right| + c$.

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

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