Question

If $n\in N$ and $I_n=\int {\left[\log \left(x\right)\right]}^{n}dx$, then$I_n+n{I}_{n-1}=$

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

Answer 1

The correct option is B

$x{\left[\log \left(x\right)\right]}^n+c$

Explanation for the correct option

Given that $I_n=\int {\left[\log \left(x\right)\right]}^{n}dx$

$\begin{array}{crcll}\therefore & I_n& =& \int {\left[\log \left(x\right)\right]}^n.1dx& \\ & & =& {\left[\log \left(x\right)\right]}^n\int 1dx-\int \left[\frac{d}{dx}{\left(\log \left(x\right)\right)}^n\int 1dx\right]dx& \left[\because \int u.v=u.\int vdx-\int \left(\frac{du}{dx}\int vdx\right)dx\right]\\ & & =& x{\left[\log \left(x\right)\right]}^n-\int \left[n{\left(\log \left(x\right)\right)}^{n-1}.\frac{1}{x}.x\right]dx& \left[\because \frac{d}{dx}f\left(g\left(x\right)\right)=f\text{'}\left(g\left(x\right)\right).g\text{'}\left(x\right);\frac{d}{dx}\log \left(x\right)=\frac{1}{x}\right]\\ & & =& x{\left[\log \left(x\right)\right]}^n-\int n{\left(\log \left(x\right)\right)}^{n-1}dx& \\ \Rightarrow & I_n& =& x{\left[\log \left(x\right)\right]}^n-n{I}_{n-1}+C& \left[\because I_n=\int {\left[\log \left(x\right)\right]}^{n}dx\right]\\ \Rightarrow & I_n+n{I}_{n-1}& =& x{\left[\log \left(x\right)\right]}^n+C& \end{array}$

Hence, the correct option is option(B) i.e. $x{\left[\log \left(x\right)\right]}^n+c$