int frac{x+1}{x(1+x e^x)} d x is equal to | 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 $e^x$: $I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)} d x$. Let $t

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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 $e^x$: $I = \int \frac{e^x(x+1)}{x e^x(1+x e^x)} d x$. Let $t = x e^x$. Then $dt = (e^x + x e^x) d x = e^x(1+x) d x$. Substituting these into the integral,we get: $I = \int \frac{dt}{t(1+t)}$. Using partial fractions: $\frac{1}{t(1+t)} = \frac{1}{t} - \frac{1}{1+t}$. Integrating both terms: $I = \int \left( \frac{1}{t} - \frac{1}{1+t} \right) dt = \log |t| - \log |1+t| + C = \log \left| \frac{t}{1+t} \right| + C$. Substituting $t = x e^x$ back: $I = \log \left| \frac{x e^x}{1+x e^x} \right| + C$.

$\int \frac{x+1}{x(1+x e^x)} d x$ is equal to

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