jee-main 2023 Q81

jee-main · India · session2_08apr_shift1 Integration by Substitution Substitution to Compute an Indefinite Integral with Initial Condition

Let $I ( x ) = \int \frac { x + 1 } { x \left( 1 + x e ^ { x } \right) ^ { 2 } } d x , x > 0$. If $\lim _ { x \rightarrow \infty } I ( x ) = 0$ then $I ( 1 )$ is equal to
(1) $\frac { e + 2 } { e + 1 } - \log _ { e } ( e + 1 )$
(2) $\frac { e + 1 } { e + 2 } + \log _ { e } ( e + 1 )$
(3) $\frac { e + 1 } { e + 2 } - \log _ { e } ( e + 1 )$
(4) $\frac { e + 2 } { e + 1 } + \log _ { e } ( e + 1 )$

Let $I ( x ) = \int \frac { x + 1 } { x \left( 1 + x e ^ { x } \right) ^ { 2 } } d x , x > 0$. If $\lim _ { x \rightarrow \infty } I ( x ) = 0$ then $I ( 1 )$ is equal to\\
(1) $\frac { e + 2 } { e + 1 } - \log _ { e } ( e + 1 )$\\
(2) $\frac { e + 1 } { e + 2 } + \log _ { e } ( e + 1 )$\\
(3) $\frac { e + 1 } { e + 2 } - \log _ { e } ( e + 1 )$\\
(4) $\frac { e + 2 } { e + 1 } + \log _ { e } ( e + 1 )$

Paper Questions

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