Let $f(x)=\int \frac{\sqrt{x}}{(1+x)^{2}} d x(x \geq 0)$. Then $f(3)-f(1)$ is equal to:

(A) $\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}$

(B) $-\frac{\pi}{12}+\frac{1}{2}+\frac{\sqrt{3}}{4}$

(C) $\frac{\pi}{6}+\frac{1}{2}-\frac{\sqrt{3}}{4}$

(D) $-\frac{\pi}{6}+\frac{1}{2}+\frac{\sqrt{3}}{4}$

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Indefinite Integrals

Let

f (x) = \int \frac{x}{( 1 + x ) ^{2}} d x (x \geq 0)

. Then

f (3) - f (1)

is equal to:

Answer: (A)

Here

f (3) - f (1) = \int_{1}^{3} \frac{x}{( 1 + x ) ^{2}} d x

. Substituting

x = t

x = t^{2}

d x = 2 t d t

, the integral becomes

\int_{1}^{3} \frac{2 t ^{2}}{( 1 + t ^{2} ) ^{2}} d t

. Integrating by parts with

u = t

and

d v = \frac{2 t d t}{( 1 + t ^{2} ) ^{2}}

, so

v = - \frac{1}{1 + t ^{2}}

, gives

[- \frac{t}{1 + t ^{2}}]_{1}^{3} + \int_{1}^{3} \frac{d t}{1 + t ^{2}} = (- \frac{3}{4} + \frac{1}{2}) + [tan^{- 1} t]_{1}^{3} = \frac{1}{2} - \frac{3}{4} + (\frac{π}{3} - \frac{π}{4}) = \frac{π}{12} + \frac{1}{2} - \frac{3}{4}

JEE Main · 2020 · 04 Sep 2020 · Shift 1 · Single correct option