Indefinite Integrals

antinerd

1. The problem statement, all variables and given/known data

Hey guys, I'm trying to teach myself how to integrate an indefinite integral.

I just am wondering what you can do with something like this:

2. Relevant equations

[tex]\int[/tex] 15/(3x+1) dx

3. The attempt at a solution

I'm trying to figure out how to go backwards, but I don't see what terms, when derived, give you 15/3x+1 dx.

Does anyone know a good way to quickly solve these sorts of problems?

Dick

What about log(3x+1)?

antinerd

Ah. But wouldn't it have to be:

15 ln(3x+1)

because its derivative would be:

15/(3x+1)

Correct?

So wouldn't the integral of the original problem be just:

[tex]\int[/tex] 15/(3x+1) dx = 15ln(3x+1)

Does it matter whether ln or log is used?

Dick

15 is wrong, do the differentiation again and don't forget the chain rule. it doesn't matter whether it's log or ln because I mean natural logarithm by both. If you want to use logs to a different base then you'll have to adjust the coefficient. log(base a)x=ln(x)/ln(a).

antinerd

OK, thanks, lemme work this out.

antinerd

Ok, what about this thing:

[tex]\int[/tex] 15/(3x+1) dx
and if I factor out the 15:
15[tex]\int[/tex] (3x+1) dx

Now, does the constant just disappear?

and the antiderivative of 1/(3x+1) is just

log (3x+1)
???

Dick

The derivative of log(3x+1) is 3/(3x+1). That's the answer now YOU tell me why.

antinerd

Because if you take the derivative as such:

log(3x+1) dx

You will get

d/dx f(g(x)) = f`(g(x))(g`(x))

Which means that:

d/dx log(3x+1) = (1/(3x+1)) (3)

= 3/(3x+1)

But so now do I have to place a constant to make the derivative 15? I'm wondering if

[tex]\int[/tex]15/(3x+1) = just log(3x+1)

Shouldn't it be 5(log(3x+1)), to give it a 15 on top?

Dick

Exactly, 5*log(3x+1).

learningphysics

I think it's safer to use ln... some people use log to refer to base-10 logarithm by default...

Also, antiderivative of 1/x = ln|x| (absolute value)

So your answer would be 5*ln|3x+1|