Question

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

• A. $\frac{n}{m+1},I_{mn-1}$
• B. $\frac{m}{n+1},I_{mn-1}$
• C. $-\frac{n}{m+1},I_{mn-1}$
• D. $\frac{n}{m+1}.I_{m-1},n-1$

Answer 1

Answer: (C)

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

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

$=(logx{)}^n\int x^m\cdot dx-\int [\frac{n(logx{)}^{n-1}}{x}\cdot \int x^m\cdot dx]dx$

$=\frac{(logx{)}^n-x^{m+1}(logx{)}^n}{m+1}\frac{n}{m+1}\int (logx{)}^{n-1}{x}^{m}dx$

$I_{m,n}=\frac{(logx{)}^n{x}^{m+1}}{m+1}\cdot (logx{)}^n-\frac{n}{m+1}{I}_{m,n-1}$

$\Rightarrow I_{m,n}-\frac{(logx{)}^n{x}^{m+1)}}{(m+1)}\cdot (logx{)}^n=\frac{-n}{m+1}{I}_{m,n+1}$