# Choices for Integrationg by Parts

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!

• 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. – Austin Mohr Mar 3 '12 at 17:09
• 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. – André Nicolas Mar 3 '12 at 18:15

## 2 Answers

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

• 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. – Oliver Spryn Mar 3 '12 at 17:16
• @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... – Mike Mar 3 '12 at 17:21
• Errummm.... that was silly mistake. Ok, that was my problem! Thank you for pointing that out Mike! How could I have missed that? – 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.

• If that is the case, then shouldn't the answer come out the be the same, regardless of what (within reason) I pick? – Oliver Spryn Mar 3 '12 at 17:17
• @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 – Manishearth Mar 3 '12 at 17:26
• 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 – Manishearth Mar 3 '12 at 17:29
• Your mistake is that your u was in the denominator. $\frac{x}{(2x+1)^2}$ makes more sense as u by LIATE. – Manishearth Mar 3 '12 at 17:35
• @spryno724 possibly. No guarantee that you'll reach an answer, though. – Manishearth Mar 3 '12 at 18:15