1
JEE Main 2023 (Online) 8th April Morning Shift
+4
-1

Let $$I(x)=\int \frac{(x+1)}{x\left(1+x e^{x}\right)^{2}} d x, x > 0$$. If $$\lim_\limits{x \rightarrow \infty} I(x)=0$$, then $$I(1)$$ is equal to :

A
$$\frac{e+1}{e+2}-\log _{e}(e+1)$$
B
$$\frac{e+1}{e+2}+\log _{e}(e+1)$$
C
$$\frac{e+2}{e+1}-\log _{e}(e+1)$$
D
$$\frac{e+2}{e+1}+\log _{e}(e+1)$$
2
JEE Main 2023 (Online) 6th April Morning Shift
+4
-1

Let $$I(x)=\int \frac{x^{2}\left(x \sec ^{2} x+\tan x\right)}{(x \tan x+1)^{2}} d x$$. If $$I(0)=0$$, then $$I\left(\frac{\pi}{4}\right)$$ is equal to :

A
$$\log _{e} \frac{(\pi+4)^{2}}{32}-\frac{\pi^{2}}{4(\pi+4)}$$
B
$$\log _{e} \frac{(\pi+4)^{2}}{16}-\frac{\pi^{2}}{4(\pi+4)}$$
C
$$\log _{e} \frac{(\pi+4)^{2}}{16}+\frac{\pi^{2}}{4(\pi+4)}$$
D
$$\log _{e} \frac{(\pi+4)^{2}}{32}+\frac{\pi^{2}}{4(\pi+4)}$$
3
JEE Main 2023 (Online) 25th January Morning Shift
+4
-1

Let $$f(x) = \int {{{2x} \over {({x^2} + 1)({x^2} + 3)}}dx}$$. If $$f(3) = {1 \over 2}({\log _e}5 - {\log _e}6)$$, then $$f(4)$$ is equal to

A
$${\log _e}19 - {\log _e}20$$
B
$${\log _e}17 - {\log _e}18$$
C
$${1 \over 2}({\log _e}19 - {\log _e}17)$$
D
$${1 \over 2}({\log _e}17 - {\log _e}19)$$
4
JEE Main 2022 (Online) 29th July Evening Shift
+4
-1

For $$I(x)=\int \frac{\sec ^{2} x-2022}{\sin ^{2022} x} d x$$, if $$I\left(\frac{\pi}{4}\right)=2^{1011}$$, then

A
$$3^{1010} I\left(\frac{\pi}{3}\right)-I\left(\frac{\pi}{6}\right)=0$$
B
$$3^{1010} I\left(\frac{\pi}{6}\right)-I\left(\frac{\pi}{3}\right)=0$$
C
$$3^{1011} I\left(\frac{\pi}{3}\right)-I\left(\frac{\pi}{6}\right)=0$$
D
$$3^{1011} I\left(\frac{\pi}{6}\right)-I\left(\frac{\pi}{3}\right)=0$$
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