8. Integration Review
≪ 7. Integration by Parts | Table of Contents | 9. Improper Integrals ≫When we learn how to take derivatives, there’s an algorithmic approach: apply the chain rule and the product rule as much as you can. We’ve seen some more sophisticated approaches involving implicit differentiation or the inverse function theorem earlier this quarter as well, but in my opinion, it’s relatively clear-cut when and how to apply them.
Contrast this with integration. Thus far, you have learned about four different techniques for computing indefinite and definite integrals:
- \(u\)-substitutions,
- trigonometric substitutions,
- integration by parts, and
- partial fraction decompositions.
Unlike with derivatives, there’s no clear-cut algorithm for which of these techniques to use. Even worse, \(u\)-substitutions and integration by parts force you to make a choice of what \(u\) (and sometimes what \(dv\)) should be; these choices are not always obvious. Additionally, there are problems where multiple techniques must be used in tandem, e.g. performing a \(u\)-substitution after integrating by parts.
Our goal is thus to identify some strategies for determining which of the above techniques ought to be used when solving an integral. Again, there’s no algorithm that can 100% determine which technique is most applicable.
I personally think through the following steps when I see an integral:
- If the integrand is a rational function…
- Try a \(u\)-substitution with \(u\) set to the denominator.
- If that doesn’t work out, perform a partial fraction decomposition.
- In each of the resulting pieces, try using the \(u\)-substitutions again, and perform trigonometric substitutions on the terms with irreducible quadratic denominators.
- If the integrand has a sum or difference of squares, possibly mixed up with squareroots, try doing a trigonometric substitution.
- In any other case, try doing a \(u\)-substitution.
- Try writing out several different choices of \(u\), even exhausting all of them, and seeing if you can get the integrand solely in terms of \(u\). There are only finitely many choices of \(u\), and you will either find one that works or find that none of them do.
- If a \(u\)-substitution doesn’t work, perform integration by parts. To determine \(u\) and \(dv\)…
- If there’s a factor that can’t be integrated (e.g. \(\arctan(x)\) or \(\ln(x+1)\)), that’s usually going to be \(u\).
- If every factor is easily integrated and differentiated, \(u\) will usually be the polynomial factor (if there is one). Otherwise, it probably doesn’t matter.
- Spam integration by parts until either you get an integral that can be directly integrated or you get the same integral you started with.
Note that each of these individual steps is meant to simplify the integral into something more manageable, and you may have to apply this thought process multiple times over to finish solving an integral.
Example 1.
Determine the indefinite integral \[\int \sqrt x \ln(x) dx.\]
Solution
Following the above steps, since the integrand is not a rational function, I should not try doing a partial fraction decomposition. Likewise, the integrand doesn’t contain a sum or difference of squares, so a trigonometric substitution is not appropriate.
An idea is to try a \(u\)-substitution. However, I only really have two choices for what this substitution would be.
- \(u= \ln(x)\) yields \(\frac{du}{dx} = \frac{1}{x}\), or \(dx = x du\). This results in the integrand being rewritten as \(\int \sqrt x \cdot u\cdot x du\), which probably can’t be rewritten in terms of \(u\) alone.
- \(u=\sqrt x\) is even worse, as there’s no shot of \(\ln x\) being rewritten in terms of \(u\), nor is there much hope that rewriting \(dx\) in terms of \(du\) will help with that.
Based on these two poor choices of substitutions, it seems that I have to use integration by parts. Since \(\ln x\) is hard to integrate, I should pick \(u = \ln x\), forcing me to pick \(dv = \sqrt x dx\). Integrating, I get \(v = \frac{2}{3}x ^{\frac{3}{2}}\), and differentiating \(u\) yields \(du = \frac{1}{x} dx.\) Together, we get \[\int \sqrt x \ln(x) dx = uv - \int v\ du = \frac{2}{3}x ^{\frac{3}{2}}\ln(x) - \int \frac{2}{3} \sqrt x dx.\] This last integral is doable using the good ol’ power rule, hence \[\int \sqrt x \ln(x) dx = \frac{2}{3}x ^{\frac{3}{2}}\ln(x) - \frac{4}{9} x ^{\frac{3}{2}} +C .\]
Example 2.
Determine \[\int \ln \left( 1+x^2 \right) dx.\]
Partial Solution
Since this integrand is again not a rational function, we should not apply a partial fraction decomposition. We do see a sum of two squares in there, though, and perhaps a trigonometric substitution could help.
Draw a triangle with legs \(1, x\) and hypotenuse \(\sqrt{1+x^2}\), marking the angle between the hypotenuse and the side of length \(1\) as \(\theta \). Then, we get \(\sec^2(\theta ) = 1+ x^2\) and \(\tan(\theta ) = x\implies \sec^2(\theta ) d \theta = dx \). Substituting this in yields \[\int \ln\left( 1+x^2 \right) dx = \int \ln \left( \sec^2 \theta \right) \sec^2(\theta ) d \theta.\] However, this is a dead end and looks harder to integrate than what we started with. For brevity’s sake, I will not demonstrate this, but neither a \(u\)-substitution nor integration by parts seems to be able to handle this monstrosity.
Going back to the drawing board, we’ve ruled out a trig substitution as a viable strategy for
\[\int \ln \left( 1+x^2 \right) dx.\]
A \(u\)-substitution won’t work either: our only real option is \(u = 1+x^2\), but since \(du = 2x\ dx\), we won’t be able to easily write our integrand in terms of \(u\) alone.
Our last option is again integrating by parts. Since \(\ln \left( 1+x^2 \right)\) can only be differentiated, we should set \(u = \ln \left( 1+x^2 \right)\), forcing \(dv = dx\). Then, \(v = x\) and \(du = \frac{2x}{1+x^2} dx\), and we get \[\int \ln \left( 1+x^2 \right) dx = x \ln \left( 1+x^2 \right) - \int \frac{2x^2}{1+x^2} dx.\] This last integral can be solved using a trigonometric substitution! Try doing that part yourself :^)
Example 3.
Determine \[\int \frac{x^5}{x^3-1}dx.\]
Solution
At first glance, this looks like a partial fraction decomposition, especially if you recognise that you can factor \(x^3-1= \left( x-1 \right)\left( x^2+x+1 \right).\) However, in this case, it’s actually beneficial to try a \(u\)-substitution for the denominator first.
Letting \(u = x^3-1\), we have \(\frac{du}{dx} = 3x^2\), or \(dx = \frac{du}{3x^2}\). Substituting these in, we have \[\int \frac{x^5}{x^3-1} dx = \int \frac{x^5}{u} \frac{du}{3x^2} = \int \frac{x^3}{3u} du.\] But hey, \(u = x^3 -1\), so \(x^3=u+1\), and we get \[\int \frac{x^5}{x^3-1} dx = \int \frac{u+1}{3u} du \int \left( \frac{1}{3}+\frac{1}{3u} \right) du = \frac{1}{3}u + \frac{1}{3}\ln \left\lvert u \right\rvert+C.\] Rewriting in terms of \(x\) yields \[\int \frac{x^5}{x^3-1}dx = \frac{1}{3} \left( x^3-1 \right) + \frac{1}{3} \ln \left\lvert x^3 - 1 \right\rvert + C. \]
In this case, performing partial fractions is still feasible, but it’s more difficult — there’s a boatload of extra work one has to do when the numerator has the same degree or higher as the denominator. But partial fractions still works and will ultimately get you to the same answer! Work it out if you feel like it.