Question

If $I_{m,n}=\int x^m(\ln x{)}^{n}dx,$ then $I_{m,n}-\frac{x^{m+1}}{m+1}(\ln x{)}^n=$
(where $m,n\in N;m,n\ge 1$)

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

Answer 1

The correct option is D $-\frac{n}{m+1}\cdot I_{m,n-1}$

Given : $I_{m,n}=\int x^m(\ln x{)}^{n}dx$
Let $u=(\ln x{)}^n,v=x^m$
$\Rightarrow I_{m,n}=\int (u\cdot v)dx$
Now applying integration by part,
$=u\cdot \int vdx-\int (\frac{du}{dx}\cdot \int vdx)dx=(\ln x{)}^n\cdot \frac{x^{m+1}}{m+1}-\int n(\ln x{)}^{n-1}\cdot \frac{1}{x}\cdot \frac{x^{m+1}}{m+1}dx=(\ln x{)}^n\cdot \frac{x^{m+1}}{m+1}-\frac{n}{m+1}\int x^m(\ln x{)}^{n-1}dx$
$\Rightarrow I_{m,n}=(\ln x{)}^n\cdot \frac{x^{m+1}}{m+1}-\frac{n}{m+1}\cdot I_{m,n-1}$
$\therefore I_{m,n}-(\ln x{)}^n\cdot \frac{x^{m+1}}{m+1}=-\frac{n}{m+1}\cdot I_{m,n-1}$