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Math Help - Indefinite integral by partial fractions

  1. #1
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    Indefinite integral by partial fractions

    Hey everyone I need some help with this problem that I have. I know that I put partial fractions into the title but I am unsure if that is truly what I need in order to complete this question. All my instructions says is:

    Solve the following integral

    This is it:

    \int\frac{{x}}{{(x-1)}{(\sqrt{x}-1)}}dx

    This is my method to solve:

    \int\frac{{x}}{{(x-1)}{(\sqrt{x}-1)}}dx = \frac{A}{x-1}+\frac{B}{\sqrt{x}-1}

    I multiplied both sides by the denominator of the left side:

    \int{x}dx = A(\sqrt{x}-1) + B(x-1)

    I factor in the A and B:

    x = A\sqrt{x} - A + Bx - B

    but this is where I am stuck at.. I mean the square root is where I am stuck at because the right and left sides aren't like terms right? Could definitely use some help with this one.

    Thanks in advance
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  2. #2
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    Re: Indefinite integral by partial fractions

    Quote Originally Posted by chalupabatman View Post
    Hey everyone I need some help with this problem that I have. I know that I put partial fractions into the title but I am unsure if that is truly what I need in order to complete this question. All my instructions says is:

    Solve the following integral

    This is it:

    \int\frac{{x}}{{(x-1)}{(\sqrt{x}-1)}}dx

    This is my method to solve:

    \int\frac{{x}}{{(x-1)}{(\sqrt{x}-1)}}dx = \frac{A}{x-1}+\frac{B}{\sqrt{x}-1}

    I multiplied both sides by the denominator of the left side:

    \int{x}dx = A(\sqrt{x}-1) + B(x-1)

    I factor in the A and B:

    x = A\sqrt{x} - A + Bx - B

    but this is where I am stuck at.. I mean the square root is where I am stuck at because the right and left sides aren't like terms right? Could definitely use some help with this one.

    Thanks in advance
    The key here is to recognize that

    $(x-1) = (\sqrt{x}-1)(\sqrt{x}+1)$ and so your denominator is actually $(\sqrt{x}-1)^2(\sqrt{x}+1)$

    Now rework it using the rules for that denominator.
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  3. #3
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    Re: Indefinite integral by partial fractions

    oh because you can technically factor the (x-1)... I didn't know that I will rework it and see what I can get
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  4. #4
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    Re: Indefinite integral by partial fractions

    I would substitute u=\sqrt{x} and then use the normal partial fractions. I think you will be able to do the partial fraction expansion using romsek's hint, but then you might have trouble integrating it.

    - Hollywood
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  5. #5
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    Re: Indefinite integral by partial fractions

    Also, do NOT include the " \int" in your calculations for the partial fractions. Writing "u" for \sqrt{x} as Hollywood suggests, that would be
    \frac{u^2}{(u- 1)^2(u+ 1)}= \frac{A}{u+1}+ \frac{B}{u- 1}+ \frac{C}{(u- 1)^2}
    Last edited by HallsofIvy; March 21st 2014 at 06:41 AM.
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  6. #6
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    Re: Indefinite integral by partial fractions

    I was thinking of actually making a u-substitution in the integral, so setting u=\sqrt{x} and du=\frac{1}{2\sqrt{x}}\,dx:

    \int\frac{x}{(x-1)(\sqrt{x}-1)}\,dx =

    \int\frac{x}{(x-1)(\sqrt{x}-1)}2\sqrt{x}\frac{1}{2\sqrt{x}}\,dx =

    \int\frac{u^2}{(u^2-1)(u-1)}2u\,du =

    \int\frac{2u^3}{(u+1)(u-1)^2}\,du

    Then expand:

    \frac{2u^3}{(u+ 1)(u- 1)^2}= \frac{A}{u+1}+ \frac{B}{u-1}+ \frac{C}{(u-1)^2}

    - Hollywood
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  7. #7
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    Re: Indefinite integral by partial fractions

    Quote Originally Posted by HallsofIvy View Post
    Also, do NOT include the " \int" in your calculations for the partial fractions. Writing "u" for \sqrt{x} as Hollywood suggests, that would be
    \frac{u^2}{(u- 1)^2(u+ 1)}= \frac{A}{u+1}+ \frac{B}{u- 1}+ \frac{C}{(u- 1)^2}
    So let me see if I got this one right.. you factored the (x-1) then you used the substitution for u = x?
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  8. #8
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    Re: Indefinite integral by partial fractions

    I tried again and used u substitution this is my new process but I still think it is wrong:

    u^{2} = x
    2udu = dx

    Then I subbed into my equation to get:

    \frac{u^{2}*2udu}{(u^{2}-1)(u-1)}

    Then I proceeded with the partial fractions:

    \frac{u^{2}*2udu}{(u^{2}-1)(u-1)} = \frac{A}{(u^{2}-1)} + \frac{B}{(u-1)}

    Multiplied both sides by the denom:

    u^{2}*2udu = A (u^{2}-1) + B (u-1)

    but when I solve this I get:

    B = 1, A = 2

    which really doesn't really make sense to me.. I will try it your way to see how I fair but I just wanted to update you guys with my work and progress
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  9. #9
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    Re: Indefinite integral by partial fractions

    Look at post #6 again. Hollywood has almost done it all for you. You didn't use the correct denominator. It's $(u-1)^2$ not $(u^2-1)$.
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  10. #10
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    Re: Indefinite integral by partial fractions

    Hey everyone sorry for the long reply I had a death in my family but I am still trying to get this one done... I did what post #6 did, but when I showed my professor he said that it was wrong because it is the same degree on the top and bottom when you factor out the denominator

     \frac{2u^{3}}{u^{3}-u^{2}-u+1}

    So he informed me to do long division which I did and I got this:

     \frac{2 * \frac{u^{2}+2u-2}{u^{3}-u^{2}-u+1}}{u^{3}-u^{2}-u+1}

    then this turned into:

     \frac{2(u^{2}+2u-2)}{(u^{3}-u^{2}-u+1)^{2}}

    which I factored to this:

    \frac{2(u^{2}+2u-2)}{(u-1)^{4}(u+1)^{2}}

    I think I did something wrong because then wouldn't I have this long partial fractions with 6 different coefficients? Like:

    \frac{2(u^{2}+2u-2)}{(u-1)^{4}(u+1)^{2}} = \frac{A}{(x+1)} + \frac{B}{(x+1)^{2}} + \frac{C}{(x-1)} \frac{D}{(x-1)^{2}} \frac{E}{(x-1)^{3}} \frac{F}{(x-1)^{4}}

    It just doesn't really make sense to me.. thanks in advance guys
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  11. #11
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    Re: Indefinite integral by partial fractions

    Your professor is correct, you need to do long division first. The denominator needs to have a higher degree than the numerator for the partial fractions expansion to work. So you need to do:

    \frac{2u^3}{u^3-u^2-u+1} = 2 + \frac{u^2+u-1}{u^3-u^2-u+1} = 2 + \frac{u^2+u-1}{(u+1)(u-1)^2}

    Now the denominator has higher degree than the numerator so you can do the partial fraction expansion:

    \frac{u^2+u-1}{(u+1)(u-1)^2} = \frac{A}{u+1}+ \frac{B}{u-1}+ \frac{C}{(u-1)^2}

    - Hollywood
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  12. #12
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    Re: Indefinite integral by partial fractions

    Quote Originally Posted by hollywood View Post
    Your professor is correct, you need to do long division first. The denominator needs to have a higher degree than the numerator for the partial fractions expansion to work. So you need to do:

    \frac{2u^3}{u^3-u^2-u+1} = 2 + \frac{u^2+u-1}{u^3-u^2-u+1} = 2 + \frac{u^2+u-1}{(u+1)(u-1)^2}

    Now the denominator has higher degree than the numerator so you can do the partial fraction expansion:

    \frac{u^2+u-1}{(u+1)(u-1)^2} = \frac{A}{u+1}+ \frac{B}{u-1}+ \frac{C}{(u-1)^2}

    - Hollywood
    BRILLIANT!! I have no idea what I was thinking hahaha.. thank you my dood I will be sure to work through this! Thanks again dood
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