Indefinite Integrals: Solving Homework Problems

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Homework Help Overview

The discussion revolves around the integration of the indefinite integral \(\int 15/(3x+1) dx\), focusing on the understanding of logarithmic functions and their derivatives in the context of integration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between integration and differentiation, questioning how to derive the integral from the given function. There is discussion about the use of logarithmic functions and the impact of constants in integration.

Discussion Status

Participants are actively engaging with the problem, offering various insights into the integration process and the use of logarithmic functions. Some have suggested specific forms of the integral, while others have pointed out the importance of the chain rule in differentiation.

Contextual Notes

There is mention of potential confusion regarding the notation of logarithms (log vs. ln) and the implications of using different bases for logarithmic functions. Participants are also considering the necessity of including constants in their solutions.

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Homework Statement



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:

Homework Equations



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

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?
 
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What about log(3x+1)?
 
Dick said:
What about log(3x+1)?

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?
 
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).
 
OK, thanks, let me work this out.
 
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)
?
 
The derivative of log(3x+1) is 3/(3x+1). That's the answer now YOU tell me why.
 
Dick said:
The derivative of log(3x+1) is 3/(3x+1). That's the answer now YOU tell me why.

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?
 
Exactly, 5*log(3x+1).
 
  • #10
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|
 

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