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Math Help - Tough Integral

  1. #1
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    Tough Integral

    I've tried everything I can think of and I'm not getting it. I'm studying for a final right now, so if there's a trick to doing this question I figure I should know it.

     \int \frac{x + 1}{\sqrt{4x-x^2}} dx

    I've tried trig substitution, integration by parts, normal substitution, etc.

    Any ideas?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Thomas View Post
    I've tried everything I can think of and I'm not getting it. I'm studying for a final right now, so if there's a trick to doing this question I figure I should know it.

     \int \frac{x + 1}{\sqrt{4x-x^2}} dx

    I've tried trig substitution, integration by parts, normal substitution, etc.

    Any ideas?
    try this

     \int \frac{x + 1}{\sqrt{4x-x^2}} dx =\int \frac{x-2}{\sqrt{4x-x^2}}dx +\int \frac{3}{\sqrt{4x-x^2}}dx

    -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}}dx+\int \frac{3}{\sqrt{4-(x-2)^2}}dx

    Now make u subs in each integrals
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    try this

     \int \frac{x + 1}{\sqrt{4x-x^2}} dx =\int \frac{x-2}{\sqrt{4x-x^2}}dx +\int \frac{3}{\sqrt{4x-x^2}}dx

    -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}}dx+\int \frac{3}{\sqrt{4-(x-2)^2}}dx

    Now make u subs in each integrals
    Yeah...but its not u-substitution for the last one it is 4arcsin\bigg(\frac{x-2}{2}\bigg)+C
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by TheEmptySet View Post
    try this

     \int \frac{x + 1}{\sqrt{4x-x^2}} dx =\int \frac{x-2}{\sqrt{4x-x^2}}dx +\int \frac{3}{\sqrt{4x-x^2}}dx

    -\frac{1}{2} \int \frac{4-2x}{\sqrt{4x-x^2}}dx+\int \frac{3}{\sqrt{4-(x-2)^2}}dx

    Now make u subs in each integrals
    Also you can make a simpler integral usign a different partial fractions decomposition
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    Yeah...but its not u-substitution for the last one it is 4arcsin\bigg(\frac{x-2}{2}\bigg)+C
    It can be 2u=x-2 du=dx

    \frac{3}{\sqrt{4-(x-2)^2}}dx=3\int \frac{1}{\sqrt{4-4u^2}}du=\frac{3}{2}\int\frac{1}{sqrt{1-u^2}}du

    Now sub u=\sin(\theta), du=\cos(\theta)

    I guess you could do it all in one step with 2\sin(\theta)=x-2

    but I usually do mine one at a time.(We all can't be inte-killer's)
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  6. #6
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    With a radical??

    Quote Originally Posted by Mathstud28 View Post
    Also you can make a simpler integral usign a different partial fractions decomposition
    I would like to see the decomp it you can please!

    I don't think it can be done.

    I thought it only worked with rational functions.
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  7. #7
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    Thanks for the help, got the answer. I don't think I've ever used, or seen, that technique before.

    But, this was supposed to be an overly hard final to make up for an easy midterm, so no wonder the questions are so hard.
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