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

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Question

(C) Let $I = \int \frac{(1 + \log x)^2}{x} \, dx$. Substitute $t = 1 + \log x$. Then,differentiating both sides with respect to $x$,we get $\frac{dt}{

Vedclass · Accepted Answer

$\frac{1}{3}(1 + \log x)^3 + c$

Vedclass · Answer

(C) Let $I = \int \frac{(1 + \log x)^2}{x} \, dx$. Substitute $t = 1 + \log x$. Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{x}$,which implies $\frac{1}{x} \, dx = dt$. Substituting these into the integral,we get $I = \int t^2 \, dt$. Integrating $t^2$ with respect to $t$,we get $I = \frac{t^3}{3} + c$. Substituting back $t = 1 + \log x$,we get $I = \frac{(1 + \log x)^3}{3} + c$.

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

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