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