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 :

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 :

Accepted Answer

Answer: C. \frac{e+2}{e+1}-\log _{e}(e+1) \begin{aligned} & \mathrm{I}=\int \frac{x+1}{x\left(1+x e^xight)^2} d x \\ & 	ext { Put } 1+x e^x=t \Rightarrow x e^x=t-1 \\ & \Rightarrow\left(x e^x+e^xight) d x=d t \\ & \Rightarrow e^x(x+1) d x=d t \end{aligned} 	herefore I=\int \frac{d t}{e^x \cdot x t^2}=\int \frac{d t}{(t-1) t^2} Let \frac{1}{t^2(t-1)}=\frac{\mathrm{A}}{(t-1)}+\frac{\mathrm{B} t+\mathrm{C}}{t^2} \Rightarrow 1=\mathrm{A} t^2+(\mathrm{B} t+\mathrm{C})(t-1) Comparing coefficients of t^2, t and constant terms, we get A+B=0, C-B=0,-C=1 On solving above equations, we get \begin{aligned} & \mathrm{C}=-1,=\mathrm{B}, \mathrm{A}=1 \\ & 	herefore \mathrm{I}=\int \frac{1}{t-1} d t+\int \frac{-t-1}{t^2} d t \\ & =\int \frac{1}{t-1} d t-\int \frac{1}{t} d t-\int \frac{1}{t^2} d t \\ & =\log |t-1|-\log |t|+\frac{1}{t}+\mathrm{C} \\ & \Rightarrow \mathrm{I}=\log \left|x e^xight|-\log \left|1+x e^xight|+\frac{1}{1+x e^x}+c \end{aligned} =\log \left|\frac{x e^x}{1+x e^x}ight|+\frac{1}{1+x e^x}+C Now, \lim _{x ightarrow \infty} \mathrm{I}(x)=0 \begin{aligned} & \Rightarrow \lim _{x ightarrow \infty}\left\{\log \left|\frac{x e^x}{1+x e^x}ight|+\frac{1}{1+x e^x}+\mathrm{C}ight\}=0 \\ & \Rightarrow \lim _{x ightarrow \infty}\left\{\log \left|\frac{e^x}{\frac{1}{x}+e^x}ight|+\frac{\frac{1}{x}}{\frac{1}{x}+e^x}+\mathrm{C}ight\} \\ & \Rightarrow 0+0+\mathrm{C}=0 \Rightarrow \mathrm{C}=0 \\ & 	herefore \mathrm{I}(x)=\log \left|\frac{x e^x}{1+x e^x}ight|+\frac{1}{1+x e^x} \\ & \Rightarrow \mathrm{I}(1)=\log \left|\frac{e}{1+e}ight|+\frac{1}{1+e}=1-\log (1+e)+\frac{1}{1+e} \\ & =\frac{2+e}{1+e}-\log |1+e| \end{aligned}

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 :

Answer

Explanation

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