The value of ∫_0^1(a-bx^2)/(≤ft(a+bx^2)^2) dx is:

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

For x ≠ 0, dividing the numerator and the denominator by \(x^2\) gives

    \( \displaystyle \frac{a-bx^2}{\left(a+bx^2\right)^2} = \frac{\frac{a}{x^2}-b}{\left(\frac{a}{x}+bx\right)^2} \)

The numerator is now, apart from its sign, the derivative of \(\frac{a}{x}+bx\). Let

    \( \displaystyle t = \frac{a}{x}+bx \)

    \( \displaystyle dt = \left(b-\frac{a}{x^2}\right)dx = -\left(\frac{a}{x^2}-b\right)dx \)

So

    \( \displaystyle \int\frac{a-bx^2}{\left(a+bx^2\right)^2}\,dx = -\int\frac{dt}{t^2} \)

    \( \displaystyle = \frac{1}{t}+C \)

    \( \displaystyle = \frac{1}{\frac{a}{x}+bx}+C \)

    \( \displaystyle = \frac{x}{a+bx^2}+C \)

The antiderivative \(\frac{x}{a+bx^2}\) is defined at x = 0 as well, so it can be used on the whole interval [0, 1]:

    \( \displaystyle \int_{0}^{1}\frac{a-bx^2}{\left(a+bx^2\right)^2}\,dx = \left[\frac{x}{a+bx^2}\right]_{0}^{1} \)

    \( \displaystyle = \frac{1}{a+b}-0 = \frac{1}{a+b} \)