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Math Help - Integration troubles

  1. #1
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    Integration troubles

    Working on a Calculus I review and two integration problems have stumped me:

    1. ∫x√(2x+3)

    2. ∫(2x-5)/(2x+2x+2)

    Help would be much appreciated because I don't know where to begin on these and have a strong belief that they will be on my quiz tomorrow.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    1. ∫x√(2x+3)


    Let u = 2x + 3

    x = (u-3)/2

    du = 2dx

    1. ∫x√(2x+3) = 1/4 integral[(u-3)sqrt(u)du]

    = 1/4 integral[u^(3/2)- 3u^(1/2)]du

    2. See attachment
    Attached Thumbnails Attached Thumbnails Integration troubles-integral.jpg  
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  3. #3
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    Ah, thank you very much. You're a scholar and a gentleman. It was the substitution that was getting me. Totally forgot that was possible. I guess that's what happens when you take Calc II three years after Calc I.
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  4. #4
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    Hello, MyHairIsOnFire!

    Here's how I'd do the first one . . .


    1)\;\;\int x\sqrt{2x+3}\,dx

    Let: u \:=\:\sqrt{2x+3}\quad\Rightarrow\quad x \:=\:\tfrac{1}{2}(u^2-3) \quad\Rightarrow\quad dx \:=\:u\,du


    Substitute: . \int \tfrac{1}{2}(u^2-3)\cdot u \cdot u\,du \;=\;\frac{1}{2}\int(u^4 - 3u^2)\,du \;=\;\frac{1}{2}\left(\frac{u^5}{5} - u^3\right) + C . =\;\frac{1}{10}\,u^3\left(u^2-5\right) + C


    Back-substitute: . \frac{1}{10}\left(\sqrt{2x+3}\right)^3\bigg[(\sqrt{2x+3})^2 - 5\bigg] + C

    . . . . . . . . . . =\;\frac{1}{10}(2x+3)^{\frac{3}{2}}\bigg[2x + 3 - 5\bigg] + C

    . . . . . . . . . . =\;\frac{1}{10}(2x+3)^{\frac{3}{2}}\bigg[2x - 2\bigg] + C

    . . . . . . . . . . =\; \frac{1}{10}(2x+3)^{\frac{3}{2}}\cdot 2(x-1) + C

    . . . . . . . . . . =\; \frac{1}{5}(2x+3)^{\frac{3}{2}}(x-1) + C

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  5. #5
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    for the second one u could do something like this:
    Numerator=A.d(Denominator)/dx+B where A and B are constants.
    once u get the constants u get 2 integrals which can be solved by standard methods. hope u get me.
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  6. #6
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    for the second integral notice that the derivative of the denominator is 4x+2. if the numerator is 4x+2, it would just become a simple u substitution of form 1/u du. so multiply the numerator by 2 (but you also have to divide by 2 and you can factor that out of the integral). the numerator becomes 4x-10. what you want is 4x+2 so add 12 to 4x-10 to get 4x-10+12 = 4x+2. but you also have to subtract 12 so you would have (4x+2)-12. now split the integral up as (4x+2)/(2x+2x+2) - 12/(2x+2x+2). use u substitution on the first integral and complete the square on the denominator of the 2nd integral.
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