Question

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

• 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

The correct option is B $x(\log x{)}^n+c$

$I_n=\int log(x{)}^n.1dx$. Applying ILATE rule taking '1' as the algebric function, we get
$I_n=x\left(log\right(x){)}^n-\int \frac{nx\left(log\right(x){)}^{n-1}}{x}dx$
$I_n=x\left(log\right(x){)}^n-\int n(log\left(x\right){)}^{n-1}dx$
$I_n=x\left(log\right(x){)}^n-n{I}_{n-1}$
$I_n+n{I}_{n-1}=x\left(log\right(x){)}^n+c$