1. ## Integration by Parts: Using 1*dx as your dv value

Not sure if you guys use a different formula (maybe different representation of variables) so I"ll put down my I B P formula:

(Integral of) u*dv = u*v - (Integral of) v*du

Now i've been looking at a couple of examples and in most of them (especially where you are going to end up with a dv value very hard to integrate) u is the entire term (not including dx of course) and dv is just 1*dx. Is this a very common/effective method?

P.S. I'm new to this forum/forums in general, how do you get the fancy math symbols on here

2. Originally Posted by someguy456345
Now i've been looking at a couple of examples and in most of them (especially where you are going to end up with a dv value very hard to integrate) u is the entire term (not including dx of course) and dv is just 1*dx. Is this a very common/effective method?
It really depends on the integrand. But yes, sometimes this is the way to go.

There is a nice rule of thumb for choosing $u$ known as the LIATE rule (though its more a guideline than a rule; there are always exceptions). Among several different types of factors, choose $u$ to be the factor that comes first in this list.

Logarithmic functions
Inverse trigonometric functions
Algebraic functions
Trigonometric functions
Exponential functions

For example, when doing $\mbox{\int\arcsin x\,dx},$ we have an inverse trigonometric factor, $\arcsin x,$ and an algebraic factor, $1.$ Using the above guideline, we let $u=\arcsin x,\;dv=dx.$

But if you were doing $\mbox{\int xe^x\,dx},$ you would prefer the algebraic factor for $u,$ leaving the exponential for $dv.$ So, $u=x,\;dv=e^x\,dx.$

3. Wow thank you so much, this is quite possibly one of the best things someone has ever shown me in 5 minutes. The LIATE rule makes perfect sense to me, as in my eyes anyway, the rule goes from hardest to easiest to integrate for subbing in U.

I really appreciate the response, thanks!