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Why is an integrated function with multiple terms...

  1. Jan 4, 2016 #1
    I have encountered this general Integral:

    "∫ 1/ax+b dx = 1/a * ln|ax+b| +C"

    I was not given a proof, but would like one, along with an easy explanation, please.
     
  2. jcsd
  3. Jan 4, 2016 #2

    jedishrfu

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    Try recasting the integral with u=ax+b and see what you get.
     
  4. Jan 4, 2016 #3
    Try to look at this from an induction kind of standpoint to see why it works.

    Otherwise, you're going to have to do this purely through symbols. Remember what terms are constant and what terms are changing.
     
  5. Jan 5, 2016 #4
    I'm afraid I don't understand what the "u" is in your equation, jedishrfu.

    And I don't understand what the connection is between induction and this formula, Thewindyfan.

    Understand that I don't presume you all are communicating worthless or nonsensical information. I suspect that there is some important and relevant information contained in your statements...but precisely what it is, and how it relates to the above equation, escapes my comprehension. Would you all please explain?
     
  6. Jan 5, 2016 #5

    Samy_A

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    Maybe we can try it with two questions:
    1) What do you get when you compute the following integral?
    $$\int \frac{1}{x} \, dx$$
    2) Do you know how to use substitutions when computing integrals?
     
  7. Jan 5, 2016 #6

    fresh_42

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    The main difficulties here might arose from the question what you regard to be a proof, i.e. what can be assumed as given. Basically the chain rule of differentiation and the fact that the natural logarithm is the anti-derivative of the function ##x → 1/x## are needed here. May we assume either of them or both or neither? Do you know what ## +C## stands for? The ##u## in question is the substitution ##u(x) := a*x + b## with ##du / dx = u'(x) = a## which leads to ##dx = du / a## and so on.
     
  8. Jan 5, 2016 #7
    Samy_A: I think that the answer to your first question is ln|x|. Am I correct in understanding however that 1/a is the coefficient of the x variable, which is during integration of x set aside and then later multiplied with the resulting integrated x, and that there is no division of the raised power of the x variable, because of the peculiar characteristics of the natural log?
    I confess that I don't know how to use substitutions when computing integrals...would you explain what this is in the above example?
     
  9. Jan 5, 2016 #8

    micromass

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    Well, I'm sorry to say this, but you will probably need to go through a calculus book and learn about substitution and integration by parts from it. No forum post can substitute that.
     
  10. Jan 5, 2016 #9
    That's okay, micromass. I've got sufficient character and determination to do the necessary work. There's no other way than through hard work, I got that. If you wanted to point me to a good online resource teaching this thing though, it would be much appreciated.
     
  11. Jan 5, 2016 #10

    Samy_A

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    More precisely,
    $$\int \frac{1}{x} \, dx \, = \ln|x| + C$$
    , where ##C## is a constant (EDIT: the constant can be different for positive and negative values of x, thanks micromass).

    The integral in your question,
    $$\int \frac{1}{ax+b} \, dx \, =\frac{1}{a} \ln|ax+b| + C$$
    , is computed from the first one using a simple substitution.

    But I agree with @micromass , basic (and important) subjects as integration methods are better learned in a book. You can use a forum such as this one if you have specific questions: members will gladly help you.
     
    Last edited: Jan 5, 2016
  12. Jan 5, 2016 #11

    micromass

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    Or more precisely: where ##C## is a locally constant function. There are more functions that differentiate to ##1/x## than just ##\ln|x|+ C## if you merely take ##C## to be a constant. Sorry for being annoying :D

    Anyway to the OP: what math are you comfortable with? Especially, what calculus do you know, and where did you learn it from?
     
  13. Jan 5, 2016 #12
    u is the common variable used in the substitution method of solving complex integrals, at least in the U.S.

    I mentioned induction in hopes that you would know what that meant, but I'm afraid you didn't understand where I was getting at. Samy_A basically summed up what I wanted to get you at: think about one simple case where you get a similar family of functions from this integral. Look at how you did that integral, and then apply it to this one here to solve it.

    My mistake for not realizing you wanted a concise proof of this solution, however. In the original post the answer seems to be miswritten with an additional "x" before you made the edit.

    edit:
    This isn't necessarily true for ALL integration problems where a substitution is efficient enough to solve the problem, but here's my view on the real basics on it. I will edit this post later with an example problem to show you this basics of u-substitution, or at least the main purpose it provides to attacking complex looking integrals.
     
  14. Jan 5, 2016 #13
    Thanks windyfan, but there's no need. It so happened that as soon as I went back to my online lessons - voila - there was u-substitution and integration by parts. Soon I'll be able to understand you all.
     
  15. Jan 5, 2016 #14

    Mark44

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    Please learn to write fractions correctly. 1/ax + b means ##\frac 1 a x + b##, which I'm sure isn't what you intended. When you write fractions with everything in one line of text, use parentheses, especially around a numerator or denominator with multiple terms.

    The integrand should be written as 1/(ax + b), or better yet, using LaTeX, as ##\frac 1 {ax + b}##. We have a LaTeX primer here -- https://www.physicsforums.com/help/latexhelp/ -- it's under the INFO menu at the top of the screen, under Help/How-to.
     
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