Question

If $I_n=\int (\ln x{)}^{n}dx,$ then $I_n+n{I}_{n-1}=$

• A. $\frac{{\left(\ln x\right)}^n}{x}+C$
• B. $x(\ln x{)}^{n-1}+C$
• C. $x(\ln x{)}^n+C$
• D. none of these

Answer 1

The correct option is C $x(\ln x{)}^n+C$

$I_n=\int (\ln x{)}^{n}dx$

$=x(\ln x{)}^n-\int \frac{x\left(n\right)(\ln x{)}^{n-1}}{x}dx$

$=x(\ln x{)}^n-n{I}_{(n-1)}$

$\Rightarrow I_n-n{I}_{n-1}=x(\ln x{)}^n$