I am working on an integration by parts problem, and, it looks like I got something incorrect somewhere. I have been told to follow LIATE (L ogarithmic, I nverse trigonometric, A lgebraic, T rigonometric, E xponential) when making a choice for selecting the u term in integration by parts. So, here is where I start:

Find $\displaystyle \int \frac{x \cdot e^{2x}}{{\left(2x + 1\right)}^{2}}$

$u = {(2x + 1)}^{2} = 4x^2 + 4x + 1$

$du = 16x + 4 \ dx$

$v = \frac{x}{2} \cdot e^{2x}$

$dv = x \cdot e^{2x} \ dx$

... and the rest of the problem is solved in this manner.

In the end I come up with:

${\left(2x + 1\right)}^{2} \cdot \frac{x}{2} \cdot e^{2x} - \left(4x^2 + x\right)\left(e^{2x}\right) + \left(16x + 1\right)\left(\frac{1}{2} \cdot e^{2x}\right) - {8e}^{2x} \ dx$

Comparing my answer with that from the student solutions handbook, here is how they start:

Find $\displaystyle \int \frac{x \cdot e^{2x}}{{\left(2x + 1\right)}^{2}}$

$u = x \cdot e^{2x}$

$du = x \cdot 2e^{2x} + e^{2x} \ dx = e^{2x} \left(2x + 1\right) \ dx$

$v = -\frac{1}{2\left(2x + 1\right)}$

$dv = \frac{1}{{\left(2x + 1\right)}^{2}} \ dx$

... and the rest of the problem is solved in this manner.

In the end, they come up with:

$$\frac{e^{2x}}{4\left(2x + 1\right)} + C$$

The answer they came up with appears to be the case since they chose their u term differently than I did. I purposefully didn't make the choice they did, since it was an exponential function, which I was taught should be the last choice for u. I chose the algebraic function.

There must be something that I am missing. Could some one please show me as to where I went wrong, and perhaps some rules that would help me make better choices for my u term in the future?

Thank you for your time!

  • $\begingroup$ In general, try to pick $u$ to be a function that has a simple derivative and $dv$ to be a function that has a simple integral. $\endgroup$ – Austin Mohr Mar 3 '12 at 17:09
  • $\begingroup$ Your calculation has errors. It should be $du=(8x+4)dx$. You had $16x+4$. You chose $dv=xe^{2x}$. But then $v$ is not what you say it is. You can check that by differentiating. The "rules" are not universal. Because of the very specific choice of $e^{2x}$, the derivative of $xe^{2x}$ happens to be $(2x+1)e^{2x}$, so there is miraculous cancellation. General rules can't take care of all very contrived special cases. $\endgroup$ – André Nicolas Mar 3 '12 at 18:15

The formula for integration by parts is $\int udv=uv-\int vdu.$ You have $\int\frac{dv}u,$ which won't work.

  • $\begingroup$ Thank Mike, but I did use v and du in the integral. Since I used integration by parts twice, maybe that is why it looks as though I had done that. $\endgroup$ – Oliver Spryn Mar 3 '12 at 17:16
  • $\begingroup$ @spryno724 What do you mean it looks like you did that? You selected a u that was in the denominator. Look at your choice of u and dv. $\int udv$ would be $\int(2x+1)^2xe^{2x}dx$, which was not what you started with... $\endgroup$ – Mike Mar 3 '12 at 17:21
  • $\begingroup$ Errummm.... that was silly mistake. Ok, that was my problem! Thank you for pointing that out Mike! How could I have missed that? $\endgroup$ – Oliver Spryn Mar 3 '12 at 17:23

There's no hard-and-fast rule for integration by parts, much like the rest of integration. You have to (unfortunately =P ) apply your mental facilities here as well.

Over here, choice of $u$ eliminated a part of the denominator. A definite plus. Again, there's no rule for this, you've got to learn to perceive the hidden things like this.

  • $\begingroup$ If that is the case, then shouldn't the answer come out the be the same, regardless of what (within reason) I pick? $\endgroup$ – Oliver Spryn Mar 3 '12 at 17:17
  • $\begingroup$ @spryno724 Integration is not like algebra. There are multiple methods of simplification, but only a few will work for a given problem. You must pass through certain 'gates' to reach it $\endgroup$ – Manishearth Mar 3 '12 at 17:26
  • $\begingroup$ Wait, is your first answer the final answer or the resultant integrand? I sort of assumed the latter as it has a dx. If not, you've made a mistake somewhere. I thought you were asking why you get something easy to integrate with one substitution, and hard with the other $\endgroup$ – Manishearth Mar 3 '12 at 17:29
  • 1
    $\begingroup$ Your mistake is that your u was in the denominator. $\frac{x}{(2x+1)^2}$ makes more sense as u by LIATE. $\endgroup$ – Manishearth Mar 3 '12 at 17:35
  • 1
    $\begingroup$ @spryno724 possibly. No guarantee that you'll reach an answer, though. $\endgroup$ – Manishearth Mar 3 '12 at 18:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.