JEE Math Practice Question
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.$$\frac{e+2}{e+1}+\log _{e}(e+1)$$
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