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Question

Mathematics Question on integral

Let I(x)=∫x+1x(1+xex)2dxI(x)=\int\frac{x+1}{x(1+xe^x)^2} dx, x>0. If lim⁡x→∞I(x)=0\lim\limits_{x\rightarrow\infin}I(x)=0, then I(1) is equal to

A

e+2e+1−loge(e+1)\frac{e+2}{e+1}-log_e(e+1)

B

e+2e+1+loge(e+1)\frac{e+2}{e+1}+log_e(e+1)

C

e+1e+2−loge(e+1)\frac{e+1}{e+2}-log_e(e+1)

D

e+1e+2+loge(e+1)\frac{e+1}{e+2}+log_e(e+1)

Answer

e+2e+1−loge(e+1)\frac{e+2}{e+1}-log_e(e+1)

Explanation

Solution

The correct option is(A): e+2e+1−loge(e+1)\frac{e+2}{e+1}-log_e(e+1)