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

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(A) Given,$I_{m, n} = \int x^m (\log x)^n \, dx$. Using integration by parts,let $u = (\log x)^n$ and $dv = x^m \, dx$. Then $du = n(\log x)^{n-1} \cd

Vedclass · Accepted Answer

$\frac{x^{m+1}}{m+1} (\log x)^n - \frac{n}{m+1} I_{m, n-1}$

Vedclass · Answer

(A) Given,$I_{m, n} = \int x^m (\log x)^n \, dx$. Using integration by parts,let $u = (\log x)^n$ and $dv = x^m \, dx$. Then $du = n(\log x)^{n-1} \cdot \frac{1}{x} \, dx$ and $v = \frac{x^{m+1}}{m+1}$. Applying the formula $\int u \, dv = uv - \int v \, du$: $I_{m, n} = (\log x)^n \cdot \frac{x^{m+1}}{m+1} - \int \frac{x^{m+1}}{m+1} \cdot n(\log x)^{n-1} \cdot \frac{1}{x} \, dx$. $I_{m, n} = \frac{x^{m+1}}{m+1} (\log x)^n - \frac{n}{m+1} \int x^m (\log x)^{n-1} \, dx$. Since $I_{m, n-1} = \int x^m (\log x)^{n-1} \, dx$,we get: $I_{m, n} = \frac{x^{m+1}}{m+1} (\log x)^n - \frac{n}{m+1} I_{m, n-1}$.

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

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