CUET UG 2024 — Mathematics Calculus

Whenever the denominator is the square of an expression, check if the integrand is the derivative of a simple fraction. Let's try differentiating $\dfrac{x}{a + bx^2}$a+bx2x​ and see what happens.

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Using the quotient rule with $u = x$u=x and $v = a + bx^2$v=a+bx2:

$$\dfrac{d}{dx}\left[\dfrac{x}{a + bx^2}\right] = \dfrac{(a + bx^2)(1) - x(2bx)}{(a + bx^2)^2}$$dxd​[a+bx2x​]=(a+bx2)2(a+bx2)(1)−x(2bx)​

$$= \dfrac{a + bx^2 - 2bx^2}{(a + bx^2)^2}$$=(a+bx2)2a+bx2−2bx2​

$$= \dfrac{a - bx^2}{(a + bx^2)^2}$$=(a+bx2)2a−bx2​

This is exactly our integrand!

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So the integral becomes:

$$\int_{0}^{1} \dfrac{a - bx^2}{(a + bx^2)^2} \, dx = \left[\dfrac{x}{a + bx^2}\right]_0^1$$∫01​(a+bx2)2a−bx2​dx=[a+bx2x​]01​

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Applying the limits:

$$= \dfrac{1}{a + b(1)^2} - \dfrac{0}{a + b(0)^2}$$=a+b(1)21​−a+b(0)20​

$$= \dfrac{1}{a + b} - 0$$=a+b1​−0

$$= \dfrac{1}{a+b}$$=a+b1​