If Im,n=∫xm(log⁡x)ndx then Im,n−xm+1(m+1)(log⁡x)n=

Im,n=∫xm(log⁡x)ndxU=(log⁡x)n.V=xm⇒Im,n=(log⁡x)nxm+1m+1-∫n(log⁡x)n-1xxm+1m+1dx⇒Im,n=(log⁡x)nxm+1m+1-nm+1∫(log⁡x)n-1xmdx⇒Im,n-(log⁡x)nxm+1m+1=-nm+1∫(log⁡x)n-1xmdx=-nm+1Im,n−1

If Im,n=∫xm(log⁡x)ndx then Im,n−xm+1(m+1)(log⁡x)n=

Im,n=∫xm(log⁡x)ndxU=(log⁡x)n.V=xm⇒Im,n=(log⁡x)nxm+1m+1-∫n(log⁡x)n-1xxm+1m+1dx⇒Im,n=(log⁡x)nxm+1m+1-nm+1∫(log⁡x)n-1xmdx⇒Im,n-(log⁡x)nxm+1m+1=-nm+1∫(log⁡x)n-1xmdx=-nm+1Im,n−1

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If $I_{m, n} = \int x^{m} (\log x)^{n} dx then I_{m, n} - \frac{x^{m + 1}}{(m + 1)} (\log x)^{n} =$

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$\frac{n}{m + 1} \cdot I_{m, n - 1}$

$\frac{m}{n + 1} \cdot I_{m, n - 1}$

$- \frac{n}{m + 1} \cdot I_{m, n - 1}$

$\frac{n}{m + 1} \cdot I_{m - 1, n - 1}$

answer is C.

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Detailed Solution

$\begin{array}{l} I_{m, n} = \int x^{m} (\log x)^{n} dx \\ U = (\log x)^{n} . V = x^{m} \\ \Rightarrow I_{m, n} = (\log x)^{n} (\frac{x^{m + 1}}{m + 1}) - \int \frac{n (\log x)^{n - 1}}{x} (\frac{x^{m + 1}}{m + 1}) dx \\ \Rightarrow I_{m, n} = (\log x)^{n} (\frac{x^{m + 1}}{m + 1}) - \frac{n}{m + 1} \int (\log x)^{n - 1} x^{m} dx \\ \Rightarrow I_{m, n} - (\log x)^{n} (\frac{x^{m + 1}}{m + 1}) = - \frac{n}{m + 1} \int (\log x)^{n - 1} x^{m} dx = - \frac{n}{m + 1} I_{m, n - 1} \end{array}$