Question

The value of $\int \frac{x^2}{(x^2+a^2)(x^2+b^2)}dx$ is

• A. $\frac{1}{b^2-a^2}\left[b{\tan }^{-1}\right]-a{\tan }^{-1}\frac{x}{a}]+C$
• B. $\frac{1}{b^2-a^2}\left[a{\tan }^{-1}\right]-b{\tan }^{-1}\frac{x}{a}]+C$
• C. $\frac{1}{b^2-a^2}\left[b{\tan }^{-1}\right]+a{\tan }^{-1}\frac{x}{a}]+C$
• D. $noneofthese$

Answer 1

The correct option is D $noneofthese$

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

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

$I_1=a^2\int \frac{dx}{(x^2+a^2)(x^2+b^2)}$

$I_1=\int \frac{a^2}{(x^2+a^2)-(x^2+b^2)}\int \frac{(x^2+a^2)-(x^2+b^2)}{(x^2+a^2)(x^2+b^2)}dx$

$I_1=\frac{a^2}{a^2-b^2}\int \frac{(x^2+a^2)dx}{(x^2+a^2)(x^2+b^2)}-\frac{a^2}{a^2-b^2}\int \frac{x^2+b^{2}dx}{(a^2+x^2)(x^2+b^2)}$

$I_1=\frac{a^2}{a^2-b^2}\int \frac{dx}{x^2-b^2}\frac{a^2}{a^2-b^2}\int \frac{dx}{x^2-a^2}$

$I_1=\frac{a^2}{a^2-b^2}\{\frac{1}{2}{\tan }^{-1}\left(\frac{x}{b}\right)-\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)\}$

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