The value of $\int_{0}^{1} \frac{a-b x^{2}}{\left(a+b x^{2}\right)^{2}} d x$ is :

Whenever the denominator is the square of an expression, check if the integrand is the derivative of a simple fraction. Let's try differentiating $\dfrac{x}{a + bx^2}$ and see what happens.

Using the quotient rule with $u = x$ and $v = a + bx^2$:

$\dfrac{d}{dx}\left[\dfrac{x}{a + bx^2}\right] = \dfrac{(a + bx^2)(1) - x(2bx)}{(a + bx^2)^2}$

$= \dfrac{a + bx^2 - 2bx^2}{(a + bx^2)^2}$

$= \dfrac{a - bx^2}{(a + bx^2)^2}$

This is exactly our integrand!

So the integral becomes:

$\int_{0}^{1} \dfrac{a - bx^2}{(a + bx^2)^2} \, dx = \left[\dfrac{x}{a + bx^2}\right]_0^1$

Applying the limits:

$= \dfrac{1}{a + b(1)^2} - \dfrac{0}{a + b(0)^2}$

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

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