# Simple integration by parts question

1. Sep 25, 2008

### protivakid

Hello,
The problem i'm working on is X225x. I know you have u = x2 and du = 2x however if dv= 25x then what is v? I know if dv were say e2x than v would be 1/2e2x but for this problem would v simply be 1/5*25x? Thank you

2. Sep 25, 2008

### SNOOTCHIEBOOCHEE

Well, there is a simple way to check if v would equal 1/5*2^5x [ it doesnt by the way] and that is take its derivitve.

when taking the deriv. of a exponential, we use this formula:

Try using that and the chain rule to come up with a better answer

3. Sep 25, 2008

### protivakid

So going by that rule v = 25xln(2) right?

4. Sep 25, 2008

Nope. If $$v=2^{5x}ln5$$, then $$dv=2^{5x}(ln5)^{2}dx$$. Then $$v$$ has to be a function so that if you take the derivative, that $$ln5$$ cancels out.

5. Sep 25, 2008

### protivakid

well if say I made u 25x would du = 5(2)4xdx? Thank you.

6. Sep 25, 2008

### protivakid

Anyone?

7. Sep 26, 2008

### HallsofIvy

Staff Emeritus
There is not much point in posting here if you are not going to pay attention to the responses. You were told that the derivative of cx is ln(c) cx. If u= 25x, then, using the chain rule, du= ln(2)25x(5x)'= ln(2)x 25x.

You might be better of with your initial idea of making dv= 25x. Since the derivative of cx is ln(c) cx, the anti-derivative of ln(c) cx is, of course, cx and, since ln(c) is just a number, the anti-derivative of cx is cx/ln(c). If dv= 25x then v= 25x/(5ln(2))+ C.

Last edited: Sep 26, 2008
8. Sep 30, 2008

### protivakid

I did get the answer and to the last poster, it's not that I am not reading the advice, it's that I don't completely understand it so I take a shot at what i think it is based on what I got from the advice and then see if I understand it correctly or if I still need help. Thank you all though.