JEE PYQ: Indefinite Integration Question 10
Question 10 - 2020 (04 Sep Shift 1)
Let $f(x) = \int \frac{\sqrt{x}}{(1+x)^2},dx$ $(x \ge 0)$. Then $f(3) - f(1)$ is equal to:
(1) $-\frac{\pi}{12} + \frac{1}{2} + \frac{\sqrt{3}}{4}$
(2) $\frac{\pi}{6} + \frac{1}{2} - \frac{\sqrt{3}}{4}$
(3) $-\frac{\pi}{6} + \frac{1}{2} + \frac{\sqrt{3}}{4}$
(4) $\frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4}$
Show Answer
Answer: (4)
Solution
Put $x = \tan^2\theta$, $dx = 2\tan\theta\sec^2\theta,d\theta$. Then $f(x) = \int 2\sin^2\theta,d\theta = \theta - \frac{\sin 2\theta}{2} + C = \tan^{-1}\sqrt{x} - \frac{\sqrt{x}}{1+x} + C$. $f(3) - f(1) = \left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right) - \left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{\pi}{12} + \frac{1}{2} - \frac{\sqrt{3}}{4}$.
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