Question

Let $I_n={\int }_0^1(x\ln x^{}{)}^{n}dx$,
If $I_4=k{\int }_2^1{x}^4\ln x^{3}dx$, then $k$ is equal to

• A. $\frac{-2}{3}$
• B. $\frac{-4}{5}$
• C. $\frac{2}{3}$
• D. $\frac{-5}{6}$

Answer 1

The correct option is B $\frac{-4}{5}$

$I_n={\int }_0^1{\left(xlnx\right)}^{n}dx$

$I_4={\int }_0^1{\left(x^{4}lnx\right)}^{4}dx$

$={\left[{\left(lnx\right)}^n\frac{x^5}{5}\right]}_0^1-{\int }_0^1(4{\left(lnx\right)}^3\times \frac{1}{x}).\left(\frac{x^5}{5}\right)dx$

$=0-\frac{4}{5}{\int }_0^1{x}^4{\left(lnx\right)}^{3}dx$

$=\frac{-4}{5}{\int }_0^1{x}^4{\left(lnx\right)}^{3}dx$

So, $k=\frac{-4}{5}$