int frac{1}{(x^2 + a^2)(x^2 + b^2)} dx = | vedclass

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(A) To evaluate the integral $I = \int \frac{1}{(x^2 + a^2)(x^2 + b^2)} dx$,we use partial fractions. We can write the integrand as: $\frac{1}{(x^2 +

Vedclass · Accepted Answer

$\frac{1}{a^2 - b^2} \left[ \frac{1}{b} 	an^{-1} \left( \frac{x}{b} ight) - \frac{1}{a} 	an^{-1} \left( \frac{x}{a} ight) ight] + c$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{1}{(x^2 + a^2)(x^2 + b^2)} dx$,we use partial fractions. We can write the integrand as: $\frac{1}{(x^2 + a^2)(x^2 + b^2)} = \frac{1}{a^2 - b^2} \left( \frac{1}{x^2 + b^2} - \frac{1}{x^2 + a^2} ight)$. Now,integrate both sides with respect to $x$: $I = \frac{1}{a^2 - b^2} \int \left( \frac{1}{x^2 + b^2} - \frac{1}{x^2 + a^2} ight) dx$. Using the standard integral formula $\int \frac{1}{x^2 + k^2} dx = \frac{1}{k} 	an^{-1} \left( \frac{x}{k} ight) + c$,we get: $I = \frac{1}{a^2 - b^2} \left[ \frac{1}{b} 	an^{-1} \left( \frac{x}{b} ight) - \frac{1}{a} 	an^{-1} \left( \frac{x}{a} ight) ight] + c$.

$\int \frac{1}{(x^2 + a^2)(x^2 + b^2)} dx = $

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