A) \[{{(x\,\log \,x)}^{n}}\]
B) \[x\,{{(\log \,x)}^{n}}\]
C) \[n\,{{(\log \,x)}^{n}}\]
D) \[{{(\log \,x)}^{n-1}}\]
Correct Answer: B
Solution :
Here we have \[{{I}_{n}}=\int{{{(\log \,x)}^{n}}dx}\] \[\therefore \] using \[\int{(u\upsilon )\,dx=[\upsilon \,.\,\,\int{\upsilon \,dx]-\int{\left\{ \frac{du}{dx}\,.\,\int{\upsilon \,dx} \right\}dx}}}\] \[={{(\log \,x)}^{n}}.\,x-\int{x\,.\,n\,{{(\log \,x)}^{n-1}}\frac{1}{x}dx}\] \[{{I}_{n}}=x\,{{(\log \,x)}^{n}}-n\,{{I}_{n-1}}\] \[\therefore \] \[{{I}_{n}}+n{{I}_{n-1}}=x\,{{(\log \,x)}^{n}}\]
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