int log x cdot [log (ex)]^{-2} dx = . . . . . | vedclass

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(A) Let $I = \int \log x \cdot [\log (ex)]^{-2} dx$ Since $\log(ex) = \log e + \log x = 1 + \log x$,the integral becomes: $I = \int \frac{\log x}{(1 +

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

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(A) Let $I = \int \log x \cdot [\log (ex)]^{-2} dx$ Since $\log(ex) = \log e + \log x = 1 + \log x$,the integral becomes: $I = \int \frac{\log x}{(1 + \log x)^{2}} dx$ Substitute $\log x = t$,which implies $x = e^{t}$ and $dx = e^{t} dt$. $I = \int \frac{t}{(1 + t)^{2}} e^{t} dt$ We can rewrite the numerator as $(t + 1 - 1)$: $I = \int e^{t} \left( \frac{t + 1 - 1}{(1 + t)^{2}} \right) dt$ $I = \int e^{t} \left( \frac{1}{1 + t} - \frac{1}{(1 + t)^{2}} \right) dt$ 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 $f'(t) = -\frac{1}{(1 + t)^{2}}$: $I = e^{t} \cdot \frac{1}{1 + t} + C$ Substituting back $t = \log x$ and $e^{t} = x$: $I = \frac{x}{1 + \log x} + C$

$\int \log x \cdot [\log (ex)]^{-2} dx = . . . . . .$

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