Answer
$$\displaystyle{\int\left(\ln x\right)^3\ dx=x(\ln x)^3-3x\left(\ln x\right)^2+6x\left(\ln x\right)-6x+C}$$
Work Step by Step
$\displaystyle{\int\left(\ln x\right)^n\ dx=x(\ln x)^n-n\int\left(\ln x\right)^{n-1}\ dx}\\
n=3\\
\displaystyle{I=(\ln x)^3\ dx}\\
\displaystyle{I=x(\ln x)^3-3\int\left(\ln x\right)^{2}\ dx}\\
\displaystyle{I=x(\ln x)^3-3\left(x\left(\ln x\right)^2-2\int\left(\ln x\right)\ dx\right)}\\
\displaystyle{I=x(\ln x)^3-3x\left(\ln x\right)^2+6\int\ln x\ dx}\\
\displaystyle{I=x(\ln x)^3-3x\left(\ln x\right)^2+6\left(x\left(\ln x\right)-1\int\left(\ln x\right)^0\ dx\right)}\\
\displaystyle{I=x(\ln x)^3-3x\left(\ln x\right)^2+6x\left(\ln x\right)-6\int1\ dx}\\
\displaystyle{I=x(\ln x)^3-3x\left(\ln x\right)^2+6x\left(\ln x\right)-6x+C}\\$
