$\int_{0}^{\infty} \frac{6}{\mathrm{e}^{3 \mathrm{x}}+6 \mathrm{e}^{2 \mathrm{x}}+11 \mathrm{e}^{\mathrm{x}}+6} \mathrm{dx}$

(B) $\log _{\mathrm{e}}\left(\frac{32}{27}\right)$

(A) $\log _{\mathrm{e}}\left(\frac{512}{81}\right)$

(C) $\log _{\mathrm{e}}\left(\frac{256}{81}\right)$

(D) $\log _{\mathrm{e}}\left(\frac{64}{27}\right)$

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Definite Integrals

\int_{0}^{\infty} \frac{6}{e ^{3 x} + 6 e ^{2 x} + 11 e ^{x} + 6} dx

Answer: (B)

The denominator factors as

(e^{x} + 1) (e^{x} + 2) (e^{x} + 3)

. Putting

t = e^{x}

so that

d x = \frac{d t}{t}

, the integral becomes

\int_{1}^{\infty} \frac{6 d t}{t ( t + 1 ) ( t + 2 ) ( t + 3 )}

. Partial fractions give

\frac{6}{t ( t + 1 ) ( t + 2 ) ( t + 3 )} = \frac{1}{t} - \frac{3}{t + 1} + \frac{3}{t + 2} - \frac{1}{t + 3}

, whose antiderivative is

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

. As

t \to \infty

this tends to

0

, and at

t = 1

it equals

ln \frac{27}{32}

. The value of the integral is therefore

0 - ln \frac{27}{32} = lo g_{e} \frac{32}{27}

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Single correct option