It's just e^x.
recall the product rule for derivatives:
Let and be two differentiable functions. By the product rule, we define,
Now, what if we integrated both sides? we get:
rearranging this equation, we obtain:
this is known as the integration by parts formula. (some books may write and for and respectively.
Now, we can apply this formula to find the anti-derivative of in the following way.
since any function can be thought of as itself times one, we think of here as . we then let and and plug those into the above formula.
You try it, and see what you come up with
...Maybe I am just not adept enough to understand this kind of thing. Maybe I should drop out of calc II and take calc I. I really just dont understand. Here is my problem, the lnx that I am trying to find the anti-derivative of is the dv already. As in, I am already doing an integration of parts where u=2x and dv=lnx. Therefore du=2 and dv=...? I dont know.
Ill start from the problem in the book. Integrate- (lnx)^2dx. So, I let u=(lnx)^2 and dv=dx. Therefore, du=2lnx and v=x. So, subing it in. I=(lnx)^2 - Integration=2xlnx. Then I let u=2x and dv=lnx. So, du=2 and dv=...?
So thats it. I suppose I am just unintelligent.
for you, then you do
it is really not as hard as it looks here, we are just pluging things into a formula after we do some calculations
now, match up the colors and/or variables, we get:
ok, so the green font is a little hard to see, but i think you can bear with it