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the fact that the derivative of x^5 is present

5x^4

not quite. But that does stand out.

So you can certainly try working with the x^4 as u, but then you'll have \(e^{ux}\)

which isn't as nice.

And you'll have \(5u/x^3\) in front because your du will be \(4x^3 dx\)

So that's a lot of mixed x's and u's.

Which often (especially in the beginning) means you're on the wrong track.

du = 5x^4 ??

u = x^5

ok good. Yes. So what is dx in terms of du ?

du = \(5x^4 dx \implies dx = ?\)

not sure I want to think I take derivative of 5x^4 ??? not sure

No, just divide.
\[du = 5x^4 dx \implies dx = \frac{1}{5x^4} du\]

so will I put e^u * 1/5x^4

Don't forget the 5x^4 you have in front of the e^u from the initial equation.

or reverse it 1/5x^4 first

Do you follow that and understand where each piece came from?

I don't really understand why 5x^4 stayed in front

Where should it have gone? It's part of the equation, I can't make it evaporate ;)

All I did was substitute u for x^5 and replaced dx with my expression with du.

But I can't do anything to the 5x^4 yet, because that's not x^5 = u.

Does that make sense?

yes it does

?? (5x^4 * 1/u *e^u) * 1/5x^4 du

never mind th 1/u should be just 1/1 shouldn't it

Umm.. close
\[\int (5x^4e^u)\frac{1}{5x^4}du = \int e^u\frac{5x^4}{5x^4}du = \int e^u du\]

I can see that because it is all multipilcation no + or -

Right.

so now will I replace u with x^5

After you integrate then you replace it back.

err take the anti-derivative.

Sorry, later on they're going to tell you that anti-derivatives are called integrals. ;p

so will the answer be e^x^5 + c

Yep.

good!!!!!
Thanks gotta go now

Though you should slap some parens in there for readability.

when you type it that is. I'm sure it's written right on your paper.

thanks