MathematicsHard64×since 2004Q4035Let I(x)=∫(x+1)x(1+xex)2dx,x>0I(x)=\int \frac{(x+1)}{x\left(1+x e^{x}\right)^{2}} d x, x > 0I(x)=∫x(1+xex)2(x+1)dx,x>0. If \lim_\limits{x \rightarrow \infty} I(x)=0, then I(1)I(1)I(1) is equal to :Ae+1e+2−loge(e+1)\frac{e+1}{e+2}-\log _{e}(e+1)e+2e+1−loge(e+1)Be+1e+2+loge(e+1)\frac{e+1}{e+2}+\log _{e}(e+1)e+2e+1+loge(e+1)Ce+2e+1−loge(e+1)\frac{e+2}{e+1}-\log _{e}(e+1)e+1e+2−loge(e+1)De+2e+1+loge(e+1)\frac{e+2}{e+1}+\log _{e}(e+1)e+1e+2+loge(e+1)Check answerSkip