Let $\mathrm{I}(\mathrm{x})=\int \frac{(\mathrm{x}+1)}{\mathrm{x}\left(1+\mathrm{xe}^{\mathrm{x}}\right)^{2}} \mathrm{dx}, \mathrm{x}>0$, If $\lim _{\mathrm{x} \rightarrow \infty} \mathrm{I}(\mathrm{x})=0$, then $\mathrm{I}(1)$ is equal to

(D) $\frac{e+2}{e+1}-\log _{e}(e+1)$

(A) $\frac{\mathrm{e}+1}{\mathrm{e}+2}-\log _{\mathrm{e}}(\mathrm{e}+1)$

(B) $\frac{e+1}{e+2}+\log _{e}(e+1)$

(C) $\frac{e+2}{e+1}+\log _{e}(e+1)$

Indefinite Integrals — JEE Main · 2023 · 08 Apr 2023 · Shift 1 Question with Solution

Let

I (x) = \int \frac{( x + 1 )}{x ( 1 + xe ^{x} ) ^{2}} dx, x > 0

,

If

lim_{x \to \infty} I (x) = 0

, then

I (1)

is equal to

Answer: (D)

Multiply numerator and denominator by

e^{x}

to get

I (x) = \int \frac{xe ^{x} + e ^{x}}{xe ^{x} ( 1 + xe ^{x} ) ^{2}} dx

. Substitute

t = 1 + x e^{x}

, so

d t = (x e^{x} + e^{x}) d x

. The integral becomes

I (x) = \int \frac{1}{( t - 1 ) t ^{2}} d t

. Using partial fractions:

\frac{1}{( t - 1 ) t ^{2}} = \frac{1}{t - 1} - \frac{1}{t} - \frac{1}{t ^{2}}

. Integrating gives

I (x) = ln ∣ t - 1∣ - ln ∣ t ∣ + \frac{1}{t} + C = \frac{1}{t} + ln \frac{t - 1}{t} + C

. Since

lim_{x \to \infty} I (x) = 0

and as

x \to \infty

t \to \infty

, we get

C = 0

. Therefore

I (1) = \frac{1}{1 + e} + ln (\frac{e}{1 + e}) = \frac{1}{1 + e} + ln e - ln (1 + e) = \frac{1}{1 + e} + 1 - ln (1 + e) = \frac{e + 2}{e + 1} - ln (e + 1)

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option