Next we look at an example of the actual application of (7.14). In the example below, thearrangement of terms is as one would work the problem with pencil and paper, except for theimplication arrows which we will omit in subsequent problems, and possibly the under-braceswhich become less and less necessary with practice.Example 7.2.2Computeintegraldisplayxexdx.Solution: First recall our boxed formula (7.14). For this problem, we writeu=xdv=exdx⇓⇑du=dxv=exintegraldisplayxbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipuprightuexdxbracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipuprightdv=uv−integraldisplayv du= (x)(ex)−integraldisplay(ex)dx=xex−ex+C.It is interesting to note that wechooseuanddv, and thencomputeduandv, with one qualifi-cation. That is thatvis not unique; the computation fromdvtovis of an antidifferentiationnature and so we really only knowvup to an additive constant. In factanyvso thatdv=exdx(we tookv=ex=⇒dv=exdx) will work in (7.14). Any additive constant, while legitimate,will eventually cancel in the final computation, and so we usually omit it. For instance, if wehad chosenv=ex+ 100, we would have haduv−integraldisplayv du=x(ex+ 100)−integraldisplay(ex+ 100)dx=xex+ 100x−ex−100x+C=xex−ex+C,with the same final answer as before.For most cases, we will just assume that the additiveconstant is zero and we will use the simplest antiderivative forv. We will also not continue towrite the implication arrows as they are technical and perhaps confusing.Now we will revisit the first example we gave in Subsection 7.2.1, using what will be ourbasic style for this method.Example 7.2.3Computeintegraldisplayxsec2x dx.Solution: First, as is standard for these, we complete a chart like in the previous example.u=xdv= sec2x dxdu=dxv= tanx
602CHAPTER 7.ADVANCED INTEGRATION TECHNIQUESintegraldisplayxsec2x dx=integraldisplayu dv=uv−integraldisplayv du=xtanx−integraldisplaytanx dx=xtanx−ln|secx|+C.Since this method is more complicated than substitution, there are more complicated consid-erations in how to apply it. First of course, one should attempt an earlier, simpler method. Butif those fail, and integration by parts is to be attempted,5the following guidelines for choosinguanddvshould be considered for our formulaintegraltextu dv=uv−integraltextv du:1.uanddvmustaccount for all factors of the original integral, and no more.1.5. Of course,dvmustcontain the differential term (for example,dx) as a factor, butcan contain more terms.2.v=integraltextdvshould be computable with relative ease.3.du=u′(x)dx(assuming the original integral was inx) should not be overly complicated.