Question

$I_{m,n}={\int }_0^1{x}^m(\log x{)}^{n}dx$, then $I_{m,n}$ is equal to

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

Answer 1

The correct option is C

-$\frac{n}{m+1}{I}_{m,n-1}$

$I_{m,n}={\int }_0^1{x}^m(\log x{)}^{n}dx={\left|\right(\log x{)}^n.\frac{x^{m+1}}{m+1}|}_0^1-{\int }_0^{1}n(\log x{)}^{n-1}.\frac{1}{x}.\frac{x^{m+1}}{m+1}dx=0-\frac{n}{m+1}{\int }_0^1{x}^m(\log x{)}^{n-1}dx=-\frac{n}{m+1}{I}_{m,n-1}$