Question

Let $I_n=\underset{0}{\overset{1}{\int }}(x\ln x{)}^{n}dx$, if $I_4=k\underset{0}{\overset{1}{\int }}{x}^4(\ln x{)}^{3}dx$, then ' $k$ ' is equal to

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

Answer 1

Answer: (B)

$I_n=\underset{0}{\overset{1}{\int }}(x\ln x{)}^{n}dx$
$I_n={\left((\ln x{)}^n\cdot \frac{x^{n+1}}{n+1}\right)|}_0^1-\frac{n}{n+1}\underset{0}{\overset{1}{\int }}{x}^n(\ln x{)}^{n-1}dx$
$\therefore I_4=\frac{-4}{5}\underset{0}{\overset{1}{\int }}{x}^4(\ln x{)}^{3}dx$