int frac{x+1}{x(1+x e^x)^2} ,d x= (where C is | vedclass

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Question

(B) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,d x$. Multiply the numerator and denominator by $e^x$: $I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)^2} \,d x$

Vedclass · Accepted Answer

$\log \left(\frac{x e^x}{1+x e^x}\right)+\frac{1}{1+x e^x}+C$

Vedclass · Answer

(B) Let $I = \int \frac{x+1}{x(1+x e^x)^2} \,d x$. Multiply the numerator and denominator by $e^x$: $I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)^2} \,d x$. Let $t = x e^x$. Then $dt = (e^x + x e^x) dx = e^x(1+x) dx$. Substituting these into the integral: $I = \int \frac{dt}{t(1+t)^2}$. Using partial fractions: $\frac{1}{t(1+t)^2} = \frac{A}{t} + \frac{B}{1+t} + \frac{C}{(1+t)^2}$. Solving for constants,we get $A=1, B=-1, C=-1$. So,$I = \int \left( \frac{1}{t} - \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) dt$. Integrating term by term: $I = \log |t| - \log |1+t| + \frac{1}{1+t} + C$. $I = \log \left| \frac{t}{1+t} \right| + \frac{1}{1+t} + C$. Substituting $t = x e^x$ back: $I = \log \left( \frac{x e^x}{1+x e^x} \right) + \frac{1}{1+x e^x} + C$.

$\int \frac{x+1}{x(1+x e^x)^2} \,d x=$ (where $C$ is a constant of integration.)

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