3.1 Integration by parts  (Page 2/5)

Begin by rewriting the integral:

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

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

Substituting into the integration-by-parts formula ( [link] ) gives

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

Substituting into [link] produces

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

Substituting back into the previous equation yields

After evaluating the last integral and simplifying, we obtain

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

Thus, we obtain

This integral appears to have only one function—namely, $\text{sin}(\text{ln}x)$ —however, we can always use the constant function 1 as the other function. In this example, let’s choose $u=\text{sin}(\text{ln}x)$ and $dv=1dx.$ (The decision to use $u=\text{sin}(\text{ln}x)$ is easy. We can’t choose $dv=\text{sin}(\text{ln}x)dx$ because if we could integrate it, we wouldn’t be using integration by parts in the first place!) Consequently, $du=(1\text{/}x)\text{cos}(\text{ln}x)dx$ and $v={{\displaystyle \int }}^{\text{}}1dx=x.$ After applying integration by parts to the integral and simplifying, we have

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 $u=\text{cos}(\text{ln}x)$ and $dv=1dx,$ making $du=\text{−}(1\text{/}x)\text{sin}(\text{ln}x)dx$ and $v={{\displaystyle \int }}^{\text{}}1dx=x.$ 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 $I={{\displaystyle \int }}^{\text{}}\text{sin}(\text{ln}x)dx.$ Thus, the equation becomes

First, add $I$ to both sides of the equation to obtain

Next, divide by 2:

Substituting $I={{\displaystyle \int }}^{\text{}}\text{sin}(\text{ln}x)dx$ again, we have

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