Question

Evaluate ${\int }_1^e{x}^n\log x.dx$

• A. $\frac{1}{(n+1{)}^2}(e^{n+1}+1)$
• B. $\frac{1}{(n+1{)}^2}(n{e}^{n+1}+1)$
• C. $\frac{1}{(n+1{)}^2}(n{e}^{n+1}-1)$
• D. $\frac{1}{(n+1{)}^2}$

Answer 1

The correct option is B $\frac{1}{(n+1{)}^2}(n{e}^{n+1}+1)$

${\int }_1^e{x}^n\log xdx$

$\int x^n\log xdx=\frac{x^{n+1}}{n+1}\log x-\frac{1}{n+1}\int x^{n}dx$

$=\log x\frac{x^{n+1}}{n+1}-\frac{1}{(n+1{)}^2}\cdot x^{n+1}+c$

${\int }_1^e{x}^n\log xdx={(\frac{x^{n+1}}{n+1}\log x-\frac{1}{(n+1{)}^2}{x}^{n+1}+c)}_1^e$

$=[\frac{e^{n+1}}{n+1}-\frac{1}{(n+1{)}^2}{e}^{n+1}+c]-(\frac{0}{n+1}-\frac{1}{(n+1{)}^2}+c)$

$=\frac{1}{(n+1{)}^2}[n{e}^{n+1}+1]$