int frac{x+1}{x(1+x e^x)} d x= | vedclass

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

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

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(B) Let $I = \int \frac{x+1}{x(1+x e^x)} d x$. Multiply the numerator and denominator by $e^x$: $I = \int \frac{(x+1) e^x}{x e^x(1+x e^x)} d x$. Let $t = 1 + x e^x$. Then $d t = (e^x + x e^x) d x = (1+x) e^x d x$. Also, from $t = 1 + x e^x$, we have $x e^x = t - 1$. Substituting these into the integral: $I = \int \frac{d t}{(t-1) t}$. Using partial fractions: $\frac{1}{(t-1) t} = \frac{1}{t-1} - \frac{1}{t}$. Integrating both sides: $I = \int \left( \frac{1}{t-1} - \frac{1}{t} \right) d t = \log |t-1| - \log |t| + C = \log \left| \frac{t-1}{t} \right| + C$. Substituting $t = 1 + x e^x$ back: $I = \log \left| \frac{x e^x}{1+x e^x} \right| + C$.

$\int \frac{x+1}{x(1+x e^x)} d x=$

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