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Math Help - Integration of x/ax^2+c

  1. #1
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    Integration of x/ax^2+c

    Hi,

    I am revising on Solids of revolution and I am stuck on integrating a particular function. I have been given examples examples of similar variations but not this one, so I have been stuck all.

    How would you integrate functions in this form:



    I have been shown when the top x is a number such as 2 or 1.

    Thanks. Sorry I can't type maths yet.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Hint: (\ln |u|)'=u'/u
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  3. #3
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    Quote Originally Posted by FernandoRevilla View Post
    Hint: (\ln |u|)'=u'/u
    Was that: ln (u) = 1/u ?

    I understand that.

    I also know that ∫1/ax+b dx = 1/a ln|ax +b|+c

    The only problem is when where there is a number there is x as in my first example.
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  4. #4
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    What is the derivative of \frac{\ln(ax^2+8)}{2a}~?
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    \int\dfrac{x\;dx}{ax^2+8}=\dfrac{1}{2a}\int\dfrac{  2ax\;dx}{ax^2+8}=\dfrac{1}{2a}\ln |ax^2+8|+C


    Edited: Sorry, I didn't see Plato's post.
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  6. #6
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    Quote Originally Posted by FernandoRevilla View Post
    \int\dfrac{x\;dx}{ax^2+8}=\dfrac{1}{2a}\int\dfrac{  2ax\;dx}{ax^2+8}=\dfrac{1}{2a}\ln |ax^2+8|+C


    Edited: Sorry, I didn't see Plato's post.
    Thanks.

    How did you come up with 1/2a? isn't it supposed to be just 1/a. What happened to the x on top?
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  7. #7
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    what is the derivative of ax^2+8?

    it's 2ax, is it not? but we only have x on top, so we put 2a inside the integral, and 1/(2a) outside.
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  8. #8
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    Quote Originally Posted by Deveno View Post
    what is the derivative of ax^2+8?

    it's 2ax, is it not? but we only have x on top, so we put 2a inside the integral, and 1/(2a) outside.
    So you're saying that when ever we have x on top we must multiply it with the derivative of the denominator at the bottom? Then multiple the current inside integral with the derivative of the denominator?
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  9. #9
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    Quote Originally Posted by FernandoRevilla View Post
    \int\dfrac{x\;dx}{ax^2+8}=\dfrac{1}{2a}\int\dfrac{  2ax\;dx}{ax^2+8}=\dfrac{1}{2a}\ln |ax^2+8|+C


    Edited: Sorry, I didn't see Plato's post.

    Sorry, I asked a stupid question previously. I now see how that is formed after revising Integration by inspection. The two is actually a factor of the differentiation of the lower part of the function. We divide that by 1 or the all expression to great rid of it when it is being differentiated.
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  10. #10
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    Perhaps you could see it better as substitution: let u= ax^2+ 8. Then du= 2a x dx. Now there are two ways to think:
    Rewrite the integral as \frac{1}{2a}\int \frac{2ax dx}{ax^2+ 8}= \frac{1}{2a}\int \frac{du}{u}

    or rewrite du= 2ax dx as \frac{1}{2a}du= dx so that \int\frac{x dx}{ax^2+ 8}= \int\frac{\frac{1}{2a}du}{u}= \frac{1}{2a}\int \frac{du}{u}
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