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Difficulty to find this integral

  1. Jun 14, 2012 #1
    Hello! I'm doing some calculus exercises and I'm having some difficulty to find the right answer.
    1. The problem statement, all variables and given/known data
    The integral is [itex]\int \frac{1}{\sqrt{5x+8}}dx[/itex] and the exercise tell me to solve this by using (a) [itex]u=5x+8[/itex] and (b) [itex]u=\sqrt{5x+8}[/itex]

    2. Relevant equations



    3. The attempt at a solution
    This is my attempt:
    (a) Well, [itex]u=5x+8[/itex] then [itex]\frac{du}{dx}=5[/itex] which means [itex]dx=\frac{du}{5}[/itex]. Then my integral is now [itex]\int \frac{du}{5 \sqrt{u}} = \frac{1}{5}ln_{\sqrt{u}} = \frac{1}{5}ln_{\sqrt{5x+8}} = \frac{1}{5}ln_{5x}ln_{8}[/itex] Is this correct?

    (b) Now [itex]u=\sqrt{5x+8}[/itex]. Then [itex]\frac{du}{dx}=\frac{1}{2\sqrt{5x+8}}[/itex] which means [itex]dx=2u[/itex] and now my integral is [itex]2 \int du = u^2 = 5x+8[/itex]. Is this correct?
    Thanks!!
     
  2. jcsd
  3. Jun 14, 2012 #2

    sharks

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    Gold Member

    (a) ##\int \frac{du}{5 \sqrt{u}}=\frac{1}{5}\int \frac{du}{ \sqrt{u}}=\frac{2\sqrt u}{5}+C##
    I get the final result from the general formula (which is very helpful if you remember it): ##\int \frac{f'(x)}{\sqrt{f(x)}}\,dx=2\sqrt{f(x)}+C##

    (b) ##\frac{du}{dx}=\frac{1}{2\sqrt{5x+8}}## is wrong.
     
  4. Jun 14, 2012 #3
    Oh, I don't know this formula. Thanks!

    About the (b), why is it wrong? This is what I did [itex]\frac{du}{dx}=\frac{d(\sqrt{5x+8})}{dx}\; \Rightarrow \frac{du}{dx}=((5x+8)^{1/2})' = \frac{1}{2}(5x+8)^{\frac{1}{2}-1}= \frac{1}{2}(5x+8)^{-\frac{1}{2}} = \frac{1}{2}\frac{1}{\sqrt{5x+8}}=\frac{1}{2\sqrt{5x+8}}[/itex]
     
  5. Jun 14, 2012 #4

    Mentallic

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    Homework Helper

    No. While [tex]\int \frac{dx}{x}=\ln(x)[/tex] you need to keep in mind that the integral only becomes a logarithm when the denominator is a linear equation in x (so of the form ax+b). We don't have that, instead you need to use the rule
    [tex]\int x^n dx=\frac{x^{n+1}}{n+1}[/tex] where n=-1/2 in this case.

    You forgot about multiplying the expression by the derivative of 5x+8=5.
    It should be [itex]dx=2u\cdot du[/itex]

    [tex]\int du= u[/tex]
     
  6. Jun 14, 2012 #5
    Wow! Lots of stupid mistakes. Now I see what I did wrong! Thank you guys very much!
     
  7. Jun 14, 2012 #6

    Ray Vickson

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    NO, NO, NO! Why would you ever think that [itex]\ln(5x+8) = \ln(5x) \ln(8)[/itex]? For that matter, what do you mean by the notation [itex]\ln_{\sqrt{u}}[/itex]? Did you mean to write
    [itex] \ln \sqrt{u}[/itex] or [itex] \ln(\sqrt{u})[/itex]?

    Anyway, your first approach, giving [tex]\frac{1}{5}\ln \sqrt{5x+8}[/tex] is wrong. The integral
    [tex] \int \frac{du}{\sqrt{u}}[/tex] is of the form
    [tex] \int u^n \, du, [/tex]
    with [itex] n \neq -1,[/itex] so can be integrated without involving logarithms.

    Your second approach is wrong: you have the wrong "du".

    BTW: to get proper typesetting of math names like log, ln, sin, cos, exp, etc. in LaTeX, you need to enter them as commands, like this: \log, \ln, \sin, \cos, \exp, etc.

    RGV
     
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