int frac{log x}{(1 + log x)^2} dx = | vedclass

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Question

(C) Let $I = \int \frac{\log x}{(1 + \log x)^2} dx$. Substitute $1 + \log x = t$,which implies $\log x = t - 1$. Differentiating both sides,we get $\f

Vedclass · Accepted Answer

$\frac{x}{1 + \log x} + c$

Vedclass · Answer

(C) Let $I = \int \frac{\log x}{(1 + \log x)^2} dx$. Substitute $1 + \log x = t$,which implies $\log x = t - 1$. Differentiating both sides,we get $\frac{1}{x} dx = dt$,so $dx = x dt$. Since $1 + \log x = t$,we have $\log x = t - 1$,so $x = e^{t-1}$. Thus,$dx = e^{t-1} dt$. Substituting these into the integral: $I = \int \frac{t - 1}{t^2} e^{t-1} dt = \int e^{t-1} \left( \frac{t}{t^2} - \frac{1}{t^2} \right) dt = \int e^{t-1} \left( \frac{1}{t} - \frac{1}{t^2} \right) dt$. Using the standard integral formula $\int e^u (f(u) + f'(u)) du = e^u f(u) + c$,where $f(t) = \frac{1}{t}$ and $f'(t) = -\frac{1}{t^2}$: $I = e^{t-1} \cdot \frac{1}{t} + c$. Substituting $t = 1 + \log x$ back: $I = \frac{e^{(1 + \log x) - 1}}{1 + \log x} + c = \frac{e^{\log x}}{1 + \log x} + c = \frac{x}{1 + \log x} + c$.

$\int \frac{\log x}{(1 + \log x)^2} dx = $

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