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Need help solving this indefinite integral via integration by substitution

  1. Jul 15, 2012 #1

    lo2

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

    Calculate the following integral:

    [itex] \int{\frac{\sqrt{x+1}}{x+5}dx} \ , x ≥ 1[/itex]

    By using the following substitution:

    [itex] t=\sqrt{x+1}[/itex]

    2. Relevant equations
    Well using the integration by substitution formula.

    3. The attempt at a solution
    So I have [itex]t=\sqrt{x+1} \Rightarrow \frac{dx}{dt}=\frac{1}{2\sqrt{x+1}} \Rightarrow dx=dt \frac{1}{2\sqrt{x+1}} [/itex]

    But I do not really have this "piece" inside the integral, so I guess you have to apply some special technique which I have forgotten about. So can someone help me?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 15, 2012 #2

    Curious3141

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    You need to put everything purely in terms of t, if possible. You can rearrange [itex]t = \sqrt{x+1}[/itex] to give you [itex]x + 5[/itex] in terms of t

    The expression for [itex]dx[/itex] can also be expressed purely in terms of t. Rewrite it this way, then substitute it back into the integral and you should find a fairly simple integral wrt t after things cancel out.
     
  4. Jul 15, 2012 #3

    HallsofIvy

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    Well, that last is dt/(2t) isn't it? Also, since [itex]t= \sqrt{x+ 1}[/itex], [itex]t^2= x+ 1[/itex] so [itex]x+ 5= t^2+ 4[/itex]

    Your integral is
    [tex]\int \frac{t}{t^2+ 4}\left(\frac{dt}{2t}\right)[/tex]

     
  5. Jul 15, 2012 #4

    lo2

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    Hey thanks a lot for the help! Well those substitutions do make sense, but my problem is that the resulting integral is as far as I can see not that easy :/

    So would I have to use integration by substitution once again to solve this, or integration by parts?

    I must admit that my integration technique is a wee bit rusty...
     
  6. Jul 15, 2012 #5

    HallsofIvy

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    What? The final integral is
    [tex]\int \frac{dt}{t^2+ 4}[/tex]
    which is a basic integral.

    (What is the derivative of arctan(x)?)
     
  7. Jul 15, 2012 #6
    If you use t=sqrt[x+1], you will get :

    I=2∫t2/(t2+4)dt . Now change the nominator, separate and integrate:


    I=2∫t2/(t2+4)dt=2∫(t2+4-4)/(t2+4)dt=

    2∫[1-4/(t2+4)]dt=2[∫1dt-4∫dt/(t2+4)]= 2 Sqrt[1 + x] - 4 ArcTan[Sqrt[1 + x]/2]
     
    Last edited: Jul 15, 2012
  8. Jul 15, 2012 #7
    So you mean derivative of arctan can be written as Sqrt[x + 1]/(x + 5)? :smile:
     
  9. Jul 15, 2012 #8

    Mark44

    Staff: Mentor

    No, HallsOfIvy is not saying that d/dx(arctan(x)) = sqrt(x+1)/(x + 5). He was giving a hint to the OP as to how to proceed.
     
  10. Jul 15, 2012 #9

    lo2

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    Ok I have come up with the following solution: (please correct me if I am wrong!)

    We have this integral:

    [itex] \int{\frac{\sqrt{x+1}}{x+5}dx} [/itex]

    Where

    [itex] t=\sqrt{x+1} \Leftrightarrow t^2+4=x+5 [/itex]

    And where

    [itex] \frac{dx}{dt}=\frac{1}{2t} \Leftrightarrow dx=\frac{1}{2t}dt [/itex]

    This gives us the following integral:

    [itex] \int{\frac{t}{t^2+4}\cdot \frac{1}{2t}dt} = \int{\frac{1}{2t^2+8}dt} = \int{\frac{1}{\frac{1}{4}t^2+1}dt} [/itex]

    The anti derivative of this is:

    [itex] Arctan{(\frac{1}{4}t)} +c = Arctan{(\frac{1}{4}\sqrt{x+1})} +c [/itex]
     
  11. Jul 15, 2012 #10

    HallsofIvy

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    Almost. The anti-derivative of [itex]1/(u^2+ 1)[/itex] is arctan(u)+ C but with
    [itex]1/(x^2/4+ 1)[/itex], u is NOT "x/4".
     
  12. Jul 15, 2012 #11

    lo2

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    Yeah ok well, I was afraid it might be so.

    But I am not sure I really can see how you are going to use the Arctan connection... So could you or someone else perhaps be very kind and explain me how to use this fact and how to rearrange the equation so that you can use this connection.

    Would be most appreciated! ;)
     
  13. Jul 15, 2012 #12
    Did you actually look at what I wrote for you? I wrote the final answer for you too.

    You have t2+4=x+5 so you get: dx/dt=2t → dx=2tdt
    so you have t2 in your nominator and not t.
     
  14. Jul 15, 2012 #13

    lo2

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    Well no I do not really understand this to be honest...

    How do you get:

    [itex] 2 \int{\frac{t^2}{t^2+4} dt}[/itex]

    From:

    [itex] \int{\frac{t}{t^2+4}\cdot \frac{1}{2t}dt} [/itex]

    As far as I can see those are not the same :/
     
  15. Jul 15, 2012 #14

    eumyang

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    Let's go back to your first attempt:
    The expressions after the first right arrow aren't correct. If
    [itex] t=\sqrt{x+1}[/itex],
    then
    [itex] dt=\frac{dx}{2\sqrt{x+1}}[/itex],
    So
    [itex] dx=2\sqrt{x+1} dt[/itex] and
    [itex] dx=2t dt[/itex]

    When you substitute into
    [itex]\int{\frac{\sqrt{x+1}}{x+5}dx}[/itex]
    you get
    [itex]\int{\frac{t}{t^2 + 4}\cdot 2t dt}[/itex], or
    [itex]\int \frac{2t^2}{t^2 + 4}[/itex]
     
  16. Jul 15, 2012 #15

    Curious3141

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    I have to apologise for my earlier post, which may have been misleading. Nothing cancels out, and you're left with:

    [tex]2\int \frac{t^2}{t^2 + 4}dt[/tex]

    as Eumyang wrote.

    To resolve that, rewrite the integrand as [itex]\frac{t^2 + 4 - 4}{t^2 + 4} = 1 - \frac{4}{t^2 + 4} = 1 - \frac{1}{{(\frac{t}{2})}^2 + 1}[/itex].

    So the integral becomes:

    [tex]2\int 1 - \frac{1}{{(\frac{t}{2})}^2 + 1} dt[/tex]

    That can be integrated either by observation, or if you want, by a simple substitution of [itex]u = \frac{t}{2}[/itex] to clarify the form.
     
  17. Jul 16, 2012 #16

    lo2

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    Ok now I have the solution solution: (please correct me if I am wrong!)

    We have this integral:

    [itex] \int{\frac{\sqrt{x+1}}{x+5}dx} [/itex]

    Where

    [itex] t=\sqrt{x+1} \Leftrightarrow t^2+4=x+5 [/itex]

    And where

    [itex] \frac{dt}{dx}=\frac{1}{2t} \Leftrightarrow dx=2tdt [/itex]

    This gives us the following integral:

    [itex] \int{\frac{t}{t^2+4}\cdot 2tdt} = 2\int{\frac{t^2}{t^2+4}dt} = 2\int{\frac{t^2+4-4}{t^2+4}dt}=2\int{1-\frac{1}{(\frac{t}{2})^2+1}dt} [/itex]

    The anti derivative of this is, where I use the substitution [itex]u=\frac12[/itex]:

    [itex] \frac{du}{dt}=\frac{1}{2} \Leftrightarrow du=\frac{1}{2}dt [/itex]


    [itex] 2\int{1-\frac{1}{(\frac{t}{2})^2+1}dt}=2\int{2(1-\frac{1}{(\frac{t}{2})^2+1}\frac12)dt}=4\int{1-\frac{1}{(u)^2+1}du}=4(u-Arctan(u))=2\sqrt{x+1}-4Arctan(\frac12 \sqrt{x+1})+c [/itex]

    So I guess this is how it should be solved?

    And is [itex] Arctan=cot^{-1} [/itex] ?
     
  18. Jul 16, 2012 #17

    Curious3141

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    A couple of typos in your post, but your solution is correct, so I believe your method is, too.

    No arctan is the inverse function of tan (tangent). It's often depicted as [itex]\tan^{-1}[/itex], and the exponent does not refer to the reciprocal, it refers to the inverse function. The cotangent (cot) is the reciprocal of tan. The inverse function of cot is called arccotangent, depicted as arccot, or [itex]\cot^{-1}[/itex], following the same convention.

    I know you're asking the question because of what Wolfram outputted. The [itex]\cot^{-1}[/itex] term in the output is referring to the arccotangent function. This is a trigonometric identity (all the -1 exponents refer to inverse functions):

    [tex]\tan^{-1}{y} = \cot^{-1}{\frac{1}{y}}[/tex]

    so you should easily be able to see that your answer is the same as Wolfram's.
     
  19. Jul 16, 2012 #18

    lo2

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    How come you are so clever?
     
  20. Jul 16, 2012 #19

    Curious3141

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    If you're talking to me, well, I'm not, really, but it's nice of you to say so. :redface:
     
  21. Jul 17, 2012 #20

    lo2

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    I was indeed referring to you! But well thanks a lot for the help :)

    Btw when solving integrals which includes division of some kind, e.g. two functions divided by each other. Is there a general approach to that, or does it depend on the concrete example?
     
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