Evaluate:
anti-deriviative
3/(x-4) dx
will I use U substitute

- anonymous

Evaluate:
anti-deriviative
3/(x-4) dx
will I use U substitute

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- anonymous

Yes.

- anonymous

Identify your composite functions f(a) and g(x) such that 3/(x-4) = f(g(x))

- anonymous

u= x-4
du= 1

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## More answers

- anonymous

Yes.

- anonymous

Well, actually not quite

- anonymous

du = 1dx

- anonymous

yes I had that just forgot to type it

- anonymous

ok good.

- anonymous

ok this is where I have trouble.....will it be 3/u ???? Doesn't look right

- anonymous

That is exactly right.
If u = x-4 then 3/(x-4) = 3/u

- anonymous

so will I add 1 to the u (exponent) to be u^2

- anonymous

Not quite. realize that you basically have \[3 * \frac{1}{u}\]
So the derivative of what function of u gives you \(\frac{1}{u}\)?

- anonymous

u^-1 ??

- anonymous

1/u is u^-1

- anonymous

So it's not that. \(\frac{d}{du}[ln\ u] = \frac{1}{u}\) remember?

- anonymous

not sure what you mean about that?

- anonymous

The derivative of ln x = 1/x right?

- anonymous

ln is the natural logarithm function.

- anonymous

oh, yes it does. Too many rules.....

- anonymous

yeah, you'll get them with practice.

- anonymous

So the anti derivative of 1/x = ln x + C

- anonymous

So 3 times the anti derivative of 1/u du = ?

- anonymous

ok lost again.....
I know we have 3/u or 3 * 1/u
would it be 3lnx + c ???

- anonymous

It would be 3(ln u) + C, but then you have to plug in what u is.

- anonymous

how about
3 ln\[\left| x-4 \right|\] + C
sorry couln't get them all in one line

- anonymous

Yep, that's it exactly.

- anonymous

I am getting there......slowly.......
Thanks I really appreciate your help.

- anonymous

Of course! Just keep practicing. You're doing very well

- anonymous

I have a test next Tue. So I will be asking questions up until then. I am sure I will talk to you again..

- anonymous

ok!

- anonymous

I got really good with the derivatives, so my brain still wants to do them.

- anonymous

Yeah, I know the feeling. Soon you'll get the hang of going the other way.

- anonymous

I will start one more question before I have to leave

- anonymous

ok

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