Answer
$$\displaystyle{\int\left(\ln x\right)^n\ dx=x(\ln x)^n-n\int\left(\ln x\right)^{n-1}\ dx}$$
Work Step by Step
$\displaystyle{I=\int\left(\ln x\right)^n\ dx}$
$\displaystyle \left[\begin{array}{ll} u=\left(\ln x\right)^n & dv=1 \\ & \\ du=n\left(\ln x\right)^{n-1}\times\frac{1}{x} & v=x \end{array}\right]$ Integration by parts
$\displaystyle{I=x(\ln x)^n-\int n\left(\ln x\right)^{n-1}\times \frac{1}{x}\times x\ dx}\\
\displaystyle{I=x(\ln x)^n-n\int\left(\ln x\right)^{n-1}\ dx}$
