int log x cdot(log x+2) dx = | vedclass

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(A) Let $I = \int \log x \cdot (\log x + 2) dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \int t(t + 2) e^t dt = \

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$x(\log x)^{2}+c$

Vedclass · Answer

(A) Let $I = \int \log x \cdot (\log x + 2) dx$. Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t dt$. Then,$I = \int t(t + 2) e^t dt = \int (t^2 + 2t) e^t dt$. Recall the standard integral formula $\int e^t [f(t) + f'(t)] dt = e^t f(t) + c$. Here,let $f(t) = t^2$,then $f'(t) = 2t$. Thus,$I = e^t (t^2) + c$. Substituting back $t = \log x$,we get $I = x(\log x)^2 + c$.

$\int \log x \cdot(\log x+2) dx =$

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