int (log x)^m x^n , dx = | vedclass

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(C) Let $I = \int (\log x)^m x^n \, dx$. Substitute $\log x = t$,which implies $x = e^t$. Then,differentiating both sides with respect to $t$,we get $

Vedclass · Accepted Answer

$\int t^m e^{(n+1)t} \, dt, x = e^t$

Vedclass · Answer

(C) Let $I = \int (\log x)^m x^n \, dx$. Substitute $\log x = t$,which implies $x = e^t$. Then,differentiating both sides with respect to $t$,we get $dx = e^t \, dt$. Substituting these into the integral: $I = \int t^m (e^t)^n \cdot e^t \, dt$ $I = \int t^m e^{nt} \cdot e^t \, dt$ $I = \int t^m e^{(n+1)t} \, dt$. Thus,the correct substitution is $x = e^t$ and the resulting integral is $\int t^m e^{(n+1)t} \, dt$.

$\int (\log x)^m x^n \, dx =$

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