Begin by rewriting the integral:

Since this integral contains the algebraic function [latex]{x}^{-3}[/latex] and the logarithmic function [latex]\text{ln}x[/latex], choose [latex]u=\text{ln}x[/latex], since L comes before A in LIATE. After we have chosen [latex]u=\text{ln}x[/latex], we must choose [latex]dv={x}^{-3}dx[/latex].

Next, since [latex]u=\text{ln}x[/latex], we have [latex]du=\frac{1}{x}dx[/latex]. Also, [latex]v={\displaystyle\int }^{\text{ }}{x}^{-3}dx=-\frac{1}{2}{x}^{-2}[/latex]. Summarizing,

Substituting into the integration-by-parts formula gives

Using LIATE, choose [latex]u={x}^{2}[/latex] and [latex]dv={e}^{3x}dx[/latex]. Thus, [latex]du=2xdx[/latex] and [latex]v=\displaystyle\int {e}^{3x}dx=\left(\frac{1}{3}\right){e}^{3x}[/latex]. Therefore,

Substituting into the integration-by-parts formula produces

We still cannot integrate [latex]\displaystyle\int \frac{2}{3}x{e}^{3x}dx[/latex] directly, but the integral now has a lower power on [latex]x[/latex]. We can evaluate this new integral by using integration by parts again. To do this, choose [latex]u=x[/latex] and [latex]dv=\frac{2}{3}{e}^{3x}dx[/latex]. Thus, [latex]du=dx[/latex] and [latex]v=\displaystyle\int \left(\frac{2}{3}\right){e}^{3x}dx=\left(\frac{2}{9}\right){e}^{3x}[/latex]. Now we have

Substituting back into the previous equation yields

After evaluating the last integral and simplifying, we obtain

Unfortunately, this process leaves us with a new integral that is very similar to the original. However, let’s see what happens when we apply integration by parts again. This time let’s choose [latex]u=\cos\left(\text{ln}x\right)[/latex] and [latex]dv=1dx[/latex], making [latex]du=\text{-}\frac{1}{x}\sin\left(\text{ln}x\right)dx[/latex] and [latex]v={\displaystyle\int }^{\text{ }}1dx=x[/latex]. Substituting, we have

After simplifying, we obtain

The last integral is now the same as the original. It may seem that we have simply gone in a circle, but now we can actually evaluate the integral. To see how to do this more clearly, substitute [latex]I={\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx[/latex]. Thus, the equation becomes

First, add [latex]I[/latex] to both sides of the equation to obtain

Next, divide by 2:

Substituting [latex]I={\displaystyle\int }^{\text{ }}\sin\left(\text{ln}x\right)dx[/latex] again, we have

From this we see that [latex]\frac{1}{2}x\sin\left(\text{ln}x\right)-\frac{1}{2}x\cos\left(\text{ln}x\right)[/latex] is an antiderivative of [latex]\sin\left(\text{ln}x\right)dx[/latex]. For the most general antiderivative, add [latex]+C\text{:}[/latex]

If this method feels a little strange at first, we can check the answer by differentiation:

If we use a strict interpretation of the mnemonic LIATE to make our choice of [latex]u[/latex], we end up with [latex]u={t}^{3}[/latex] and [latex]dv={e}^{{t}^{2}}dt[/latex]. Unfortunately, this choice won’t work because we are unable to evaluate [latex]{\displaystyle\int }^{\text{ }}{e}^{{t}^{2}}dt[/latex]. However, since we can evaluate [latex]{\displaystyle\int }^{\text{ }}t{e}^{{t}^{2}}dx[/latex], we can try choosing [latex]u={t}^{2}[/latex] and [latex]dv=t{e}^{{t}^{2}}dt[/latex]. With these choices we have

Thus, we obtain