Question

If $n \in N$ and $I_n =\int (\log x)^n dx$ , then $I_n+nI_{n-1}$ 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 , $I_{n}=\int\left(log\,x\right)^{n}\,dx$
On initegrating by parts, we get
$I_{n}=\left(log\,x\right)^{n}\cdot x-\int x\cdot n\left(log\,x\right)^{n-1} \frac{1}{x} \,dx$
$I_{n}=x\left(log\, x\right)^{n}-nI_{n-1}$
$\therefore I_{n}+nI_{n-1}=x\left(log\,x\right)^{n}+C$