The value of $\int\limits_0^{1} \frac{a-b x^2}{\left(a+b x^2\right)^2} d x$ is:

$\frac{1}{a+b}$

The correct answer is Option (4) → $\frac{1}{a+b}$

$\int\limits_0^{1} \frac{a-b x^2}{\left(a+b x^2\right)^2} d x$

$I=\int\limits_0^{1}\frac{\frac{a}{x^2}-b}{\left(\frac{a}{x}+bx\right)^2}dx$

let $y=\frac{a}{x}+bx$

$dy=-\frac{a}{x^2}+bdx$

$x→0,y→∞$

$x→1,y→a+b$

$I=\int\limits_{∞}^{a+b}-\frac{dy}{y^2}=\int\limits^{∞}_{a+b}\frac{dy}{y^2}$

$=\left[\frac{1}{y}\right]_{∞}^{a+b}$

$=\frac{1}{a+b}-0$

$=\frac{1}{a+b}$