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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

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)} = $

Similar Questions

The value of $\int \frac{dx}{x(x^{n}+1)}$ is

Integrate the rational function: $\frac{2}{(1-x)(1+x^{2})}$

$\int \frac{x}{x^3-3 x+2} d x=$

$\int \frac{\tan x}{\cos x(\sec x-1)(\sec x-2)} d x=$ . . . . . . $+c$

$\int \frac{x^{2}}{(x+1)(x+2)^{2}} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module