Question

Use whatever method you like to solve each of the given definite and indefinite integrals. These integrals are neither in order of difficulty nor in order of technique. Many of the integrals can be solved in more than one way.

$\int {\left(\ln x\right)}^{3}dx$

Short answer

The result is$\begin{array}{rcl}& & x{\left(\ln x\right)}^3-3x{\left(\ln x\right)}^2+6x\ln x-6x+C\end{array}$.

Step-by-step solution

Step 1. Given information.

Consider the given information.

$\int {\left(\ln x\right)}^{3}dx$

Step 2. Find the integral.

Apply the by part's method to solve the integral.

$\begin{array}{rcl}\int {\left(\ln x\right)}^{3}dx& =& x{\left(\ln x\right)}^3-\int x3{\left(\ln x\right)}^{2}dx\\ & =& x{\left(\ln x\right)}^3-\int x·3{\left(\ln x\right)}^2·\frac{1}{x}dx\\ & =& x{\left(\ln x\right)}^3-3\int {\left(\ln x\right)}^{2}dx\\ & =& x{\left(\ln x\right)}^3-3\left[x{\left(\ln x\right)}^2-2\int \ln xdx\right]\\ & =& x{\left(\ln x\right)}^3-3\left[x{\left(\ln x\right)}^2-2\left[x\ln x-x\right]\right]+C\\ & =& x{\left(\ln x\right)}^3-\left[3x{\left(\ln x\right)}^2-6x\ln x+6x\right]+C\\ & =& x{\left(\ln x\right)}^3-3x{\left(\ln x\right)}^2+6x\ln x-6x+C\end{array}$