Simple integration by parts question

[protivakid]

Hello,
The problem I'm working on is X²2^5x. I know you have u = x² and du = 2x however if dv= 2^5x then what is v? I know if dv were say e^2x than v would be 1/2e^2x but for this problem would v simply be 1/5*2^5x? Thank you

 

[SNOOTCHIEBOOCHEE]

Well, there is a simple way to check if v would equal 1/5*2^5x [ it doesn't 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

 

[protivakid]

So going by that rule v = 2^5xln(2) right?

 

[adartsesirhc]

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.

 

[protivakid]

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

 

[protivakid]

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

Anyone?

 

[HallsofIvy]

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 c^x is ln(c) c^x. If u= 2^5x, then, using the chain rule, du= ln(2)2^5x(5x)'= ln(2)x 2^5x.

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

 

[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.