JEE Main 2020 (Online) 4th September Morning ShiftQuestion 1
Let $f(x)=\int\frac{\sqrt x}{(1+x)^2}dx$ ($x\ge0$). Then $f(3)-f(1)$ is equal to :
Indefinite Integration
Solution
Concept. Put $x=s^2$ to remove the root; then split the rational function. Formulas used. (1) $\int\dfrac{ds}{1+s^2}=\tan^{-1}s$(2) $\dfrac{2s^2}{(1+s^2)^2}=\dfrac{2}{1+s^2}-\dfrac{2}{(1+s^2)^2}$ and $\int\dfrac{ds}{(1+s^2)^2}=\dfrac12\left(\tan^{-1}s+\dfrac{s}{1+s^2}\right)$Step 1 - Substitute. $x=s^2$, $dx=2s\,ds$, $s\ge0$: $$f=\int\frac{s\cdot2s}{(1+s^2)^2}\,ds=\int\frac{2s^2}{(1+s^2)^2}\,ds.$$ Step 2 - Integrate. By (1) and (2), $$f=2\tan^{-1}s-\left(\tan^{-1}s+\frac{s}{1+s^2}\right)=\tan^{-1}s-\frac{s}{1+s^2}.$$ Step 3 - Values. $x=3$ means $s=\sqrt3$; $x=1$ means $s=1$: $$f(3)-f(1)=\left(\frac\pi3-\frac{\sqrt3}{4}\right)-\left(\frac\pi4-\frac12\right)=\frac{\pi}{12}+\frac12-\frac{\sqrt3}{4}.$$ Answer: (B)