Let I(x)= (x+1)x(1+xe^x)^2dx,x0.

Question

Let $I(x)=\int \frac{(x+1)}{x{(1+x{e}^{x})}^{2}}dx,x>0$. If $\underset{x	o \infty}{lim}I(x)=0$, then $I(1)$ is equal to :

Accepted Answer

$\begin{gathered} I=\int \frac{x+1}{x{(1+x{e}^{x})}^{2}}dx \ 1+x{e}^{x}=t\Rightarrow x{e}^{x}=t-1 \ \Rightarrow (x{e}^{x}+{e}^{x})dx=dt \ \Rightarrow {e}^{x}(x+1)dx=dt \end{gathered}$

$	herefore I=\int \frac{dt}{{e}^{x}\cdot x{t}^{2}}=\int \frac{dt}{(t-1){t}^{2}}$

Let $\frac{1}{{t}^{2}(t-1)}=\frac{A}{(t-1)}+\frac{Bt+C}{{t}^{2}}$

$\Rightarrow 1=A{t}^{2}+(Bt+C)(t-1)$

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

$\begin{gathered} C=-1,=B,A=1 \ 	herefore I=\int \frac{1}{t-1}dt+\int \frac{-t-1}{{t}^{2}}dt \ =\int \frac{1}{t-1}dt-\int \frac{1}{t}dt-\int \frac{1}{{t}^{2}}dt \ =\log |t-1|-\log |t|+\frac{1}{t}+C \ \Rightarrow I=\log |x{e}^{x}|-\log |1+x{e}^{x}|+\frac{1}{1+x{e}^{x}}+c \end{gathered}$

$=\log |\frac{x{e}^{x}}{1+x{e}^{x}}|+\frac{1}{1+x{e}^{x}}+C$

Now, $\underset{x	o \infty}{lim}I(x)=0$

$\begin{gathered} \Rightarrow \underset{x	o \infty}{lim}{\log |\frac{x{e}^{x}}{1+x{e}^{x}}|+\frac{1}{1+x{e}^{x}}+C}=0 \ \Rightarrow \underset{x	o \infty}{lim}{\log |\frac{{e}^{x}}{\frac{1}{x}+{e}^{x}}|+\frac{\frac{1}{x}}{\frac{1}{x}+{e}^{x}}+C} \ \Rightarrow 0+0+C=0\Rightarrow C=0 \ 	herefore I(x)=\log |\frac{x{e}^{x}}{1+x{e}^{x}}|+\frac{1}{1+x{e}^{x}} \ \Rightarrow I(1)=\log |\frac{e}{1+e}|+\frac{1}{1+e}=1-\log (1+e)+\frac{1}{1+e} \ =\frac{2+e}{1+e}-\log |1+e| \end{gathered}$

Solutions · 1