Question

If $I_n={\int }_1^{e^2}\left(lnx{)}^{n}d\right(x^2)=I_8+4I_7=$

• A. 1
• B. e
• C. $e^2$
• D. $\frac{e^2-1}{2}$

Answer 1

The correct option is C $e^2$

$I_n={\int }_1^{e^2}\left(lnx{)}^{n}d\right(x^2)={\int }_1^{e}2x(lnx{)}^{n}dx{\begin{array}{c}{I}_n=2(lnx{)}^n\frac{x^2}{2}\end{array}|}_1^e-2{\int }_1^e\frac{n(lnx{)}^{n-1}}{x}.\frac{x^2}{2}dx=e^2-\frac{n}{2}{I}_{n-1}\Rightarrow I_n+\frac{n}{2}{I}_{n-1}=e^2\Rightarrow I_8=4I_7=e^2$