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(C) Let $I = \int \frac{\log _e x}{(1+\log _e x)^2} dx$. Substitute $\log _e x = t$,which implies $x = e^t$ and $dx = e^t dt$. Substituting these into

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

Vedclass · Answer

(C) Let $I = \int \frac{\log _e x}{(1+\log _e x)^2} dx$. Substitute $\log _e x = t$,which implies $x = e^t$ and $dx = e^t dt$. Substituting these into the integral,we get: $I = \int \frac{t}{(1+t)^2} e^t dt$. We can rewrite the numerator as $(t+1-1)$: $I = \int \frac{t+1-1}{(1+t)^2} e^t dt = \int \left( \frac{1}{1+t} - \frac{1}{(1+t)^2} \right) e^t dt$. Using the standard integral formula $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$,where $f(t) = \frac{1}{1+t}$ and $f'(t) = -\frac{1}{(1+t)^2}$: $I = e^t \left( \frac{1}{1+t} \right) + C$. Substituting $t = \log _e x$ back into the equation: $I = e^{\log _e x} \left( \frac{1}{1+\log _e x} \right) + C = \frac{x}{1+\log _e x} + C$.

$\int \frac{\log _e x}{\left(1+\log _e x\right)^2} d x=$

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