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Question

(C) To evaluate the integral $I = \int (\log x)^2 \, dx$,we use the method of substitution.Let $\log x = t$,which implies $x = e^t$. Therefore,$dx = e

Vedclass · Accepted Answer

$x(\log x)^2 - 2x\log x + 2x + c$

Vedclass · Answer

(C) To evaluate the integral $I = \int (\log x)^2 \, dx$,we use the method of substitution.Let $\log x = t$,which implies $x = e^t$. Therefore,$dx = e^t \, dt$.Substituting these into the integral,we get:$I = \int t^2 e^t \, dt$.Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t^2$ and $dv = e^t \, dt$:$du = 2t \, dt$ and $v = e^t$.$I = t^2 e^t - \int 2t e^t \, dt = t^2 e^t - 2 \int t e^t \, dt$.Applying integration by parts again for $\int t e^t \, dt$:$I = t^2 e^t - 2(t e^t - \int e^t \, dt) = t^2 e^t - 2t e^t + 2e^t + c$.Substituting $t = \log x$ and $e^t = x$ back into the expression:$I = x(\log x)^2 - 2x \log x + 2x + c$.

$\int (\log x)^2 \, dx = $

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