Question

$\int \frac{(1+\log x{)}^n}{x}dx=$

• A. $(1+\log x{)}^n+c$
• B. $\frac{(1+\log x{)}^{n+1}}{n+1}+c$
• C. $-\frac{(1+\log x{)}^{n+1}}{n+1}+c$
• D. $\frac{(1+\log x{)}^n}{n+1}+c$

Answer 1

Answer: (B)

$\frac{(1+\log x{)}^{n+1}}{n+1}+c$

$\int {\frac{(1+logx)}{x}}^{n}dx.$
$\frac{1}{x}.dx=dlogx$
$logx=t$
$\Rightarrow 1/xdx=dt$
$\int (1+t{)}^{n}dt=\frac{(1+t{)}^{n+1}}{n+1}+c$
$=\frac{(1+logx{)}^{n+1}}{n+1}+c$