Question

lf $I_n=\int (\log x{)}^{n}dx$, then $I_6+6I_5=$

• A. $x(\log x{)}^5$
• B. $-x(\log x{)}^5$
• C. $x(\log x{)}^6$
• D. $-x(\log x{)}^6$

Answer 1

Answer: (C)

$x(\log x{)}^6$

$I_n=\int (logx{)}^{n}dx$

$I_n=\int (logx{)}^n\frac{1}{x}dx$

$=(logx{)}^n\int 1.dx-[\int \left(n\right)(logx{)}^{n-1}-\frac{1}{x}x]dx$

$I_n=(logx{)}^n-x-n\int (logx{)}^{n-1}dx$

Let n=6

$I_6=(logx{)}^{6}x-6\int (logx{)}^{5}dx$

$=I_6+6I_5=x(logx{)}^6+c$

$\Rightarrow I_6+6I_5=x(logx{)}^6+c$