Question

Let $I_n={\int }_0^1(xlnx{)}^{n}dx$, if $I_4=k{\int }_0^1{x}^4(lnx{)}^{3}dx$, then k is

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

Answer 1

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

$I_4={\int }_0^1(x\ln x{)}^{4}dx$

$={\int }_0^1{x}^4(\ln x{)}^{4}dx$

$=(\ln x{)}^4\frac{x^5}{5}-{\int }_0^{1}4(\ln x{)}^3\frac{1}{x}.\frac{x^5}{5}dx$

$I_4=(\ln x{)}^4\frac{x^5}{5}-\frac{4}{5}{\int }_0^1(\ln x{)}^3{x}^{4}dx$

$K=\frac{-4}{5}$