The value of the integer $\int_{0}^{1} \frac{\sqrt{x} d x}{(1+x)(1+3 x)(3+x)}$ is:

(B) $\frac{\pi}{8}\left(1-\frac{\sqrt{3}}{2}\right)$

(A) $\frac{\pi}{4}\left(1-\frac{\sqrt{3}}{6}\right)$

(C) $\frac{\pi}{4}\left(1-\frac{\sqrt{3}}{2}\right)$

(D) $\frac{\pi}{8}\left(1-\frac{\sqrt{3}}{6}\right)$

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Definite Integrals

The value of the integer

\int_{0}^{1} \frac{x d x}{( 1 + x ) ( 1 + 3 x ) ( 3 + x )}

is:

Answer: (B)

Put

x = t^{2}

d x = 2 t d t

, so

I = 2 \int_{0}^{1} \frac{t ^{2} d t}{( 1 + t ^{2} ) ( 1 + 3 t ^{2} ) ( 3 + t ^{2} )} .

Partial fractions give

\frac{t ^{2}}{( 1 + t ^{2} ) ( 1 + 3 t ^{2} ) ( 3 + t ^{2} )} = \frac{1}{4 ( 1 + t ^{2} )} - \frac{3}{16 ( 1 + 3 t ^{2} )} - \frac{3}{16 ( 3 + t ^{2} )}

. With

\int_{0}^{1} \frac{d t}{1 + t ^{2}} = \frac{π}{4}

\int_{0}^{1} \frac{d t}{1 + 3 t ^{2}} = \frac{π}{3 3}

and

\int_{0}^{1} \frac{d t}{3 + t ^{2}} = \frac{π}{6 3}

I = 2 (\frac{π}{16} - \frac{π}{16 3} - \frac{π}{32 3}) = \frac{π}{8} - \frac{3 π}{16} = \frac{π}{8} (1 - \frac{3}{2}) .

The value of the integral is

\frac{π}{8} (1 - \frac{3}{2}) .

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option