Let I(x) = int frac{(x+1)}{x(1+x e^x)^2} dx, | vedclass

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(D) Given $I(x) = \int \frac{x+1}{x(1+x e^x)^2} dx$. Multiply numerator and denominator by $e^x$: $I(x) = \int \frac{(x+1)e^x}{x e^x(1+x e^x)^2} dx$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given $I(x) = \int \frac{x+1}{x(1+x e^x)^2} dx$. Multiply numerator and denominator by $e^x$: $I(x) = \int \frac{(x+1)e^x}{x e^x(1+x e^x)^2} dx$. Let $u = x e^x$,then $du = (e^x + x e^x) dx = e^x(1+x) dx$. Substituting this into the integral: $I(x) = \int \frac{du}{u(1+u)^2}$. Using partial fractions: $\frac{1}{u(1+u)^2} = \frac{A}{u} + \frac{B}{1+u} + \frac{C}{(1+u)^2}$. $1 = A(1+u)^2 + Bu(1+u) + Cu$. For $u=0$,$A=1$. For $u=-1$,$C=-1$. Comparing coefficients of $u^2$: $A+B=0 \implies B=-1$. So,$I(x) = \int (\frac{1}{u} - \frac{1}{1+u} - \frac{1}{(1+u)^2}) du = \log|u| - \log|1+u| + \frac{1}{1+u} + C$. $I(x) = \log|\frac{u}{1+u}| + \frac{1}{1+u} + C = \log|\frac{x e^x}{1+x e^x}| + \frac{1}{1+x e^x} + C$. As $x \rightarrow \infty$,$\frac{x e^x}{1+x e^x} \rightarrow 1$,so $\log(1) = 0$ and $\frac{1}{1+x e^x} \rightarrow 0$. Thus,$\lim_{x \rightarrow \infty} I(x) = 0 + 0 + C = 0 \implies C = 0$. $I(1) = \log(\frac{e}{1+e}) + \frac{1}{1+e} = \log(e) - \log(1+e) + \frac{1}{1+e} = 1 - \log(1+e) + \frac{1}{1+e} = \frac{1+e+1}{1+e} - \log(1+e) = \frac{e+2}{e+1} - \log_e(e+1)$.

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

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