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

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(A) Let $I = \int {\log x(\log x + 2) \, dx}$.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 $I = \int {\log x(\log x + 2) \, dx}$.Substitute $\log x = t$,which implies $x = e^t$ and $dx = e^t \, dt$.Then,$I = \int t(t + 2) e^t \, dt = \int (t^2 + 2t) e^t \, dt$.We know that $\int e^t [f(t) + f'(t)] \, dt = e^t f(t) + c$.Here,$f(t) = t^2$ and $f'(t) = 2t$.Therefore,$I = e^t \cdot t^2 + c$.Substituting back $t = \log x$ and $e^t = x$,we get $I = x{(\log x)^2} + c$.

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

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