Question

If $n\in N$ and $I_n=\int (\log x{)}^{n}dx$ , then $I_n+n{I}_{n-1}$ is equal to

• A. $\frac{(\log x{)}^{n+1}}{n+1}$
• B. $x(\log x{)}^n+c$
• C. $(\log x{)}^{n-1}$
• D. $\frac{(\log x{)}^n}{n}$

Answer 1

Answer: (B)

Here , $I_n=\int {\left(logx\right)}^{n}dx$
On initegrating by parts, we get
$I_n={\left(logx\right)}^n\cdot x-\int x\cdot n{\left(logx\right)}^{n-1}\frac{1}{x}dx$
$I_n=x{\left(logx\right)}^n-n{I}_{n-1}$
$\therefore I_n+n{I}_{n-1}=x{\left(logx\right)}^n+C$