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Math Help - Am I Wrong (part 2)?

  1. #1
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    Am I Wrong (part 2)?

    Problem 15: \int \sqrt{12 + 4x^2}

    I get:

    x\sqrt{x^2 + 3}+ 3 \ln(\frac{\sqrt{x^2 + 3} + x}{\sqrt{3}})

    the book has the same answer except the radical in the denominator of the natural log part is gone. Where did I go wrong?

    Thanks
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  2. #2
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    What substitution did you make?

    I would choose \displaystyle x = \sqrt{3}\tan{u}

    \displaystyle \int \sqrt{12+4x^2}~dx = \sqrt{x^2+3}+3\sinh^{-1}\left(\frac{x}{\sqrt{3}}\right)+C
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  3. #3
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    Thats an indefinite integral so your answer should be x\sqrt{x^2+ 3}+ 3ln(\frac{\sqrt{x^2+ 3}+ x}{\sqrt{3}})+ C which is the same as x\sqrt{x^2+ 3}+ 3ln(\sqrt{x^2+ 3}+ x)- 3ln(\sqrt{3})+ C which is the same as x\sqrt{x^2+ 3}+ 3ln(\sqrt{x^2+ 3}+ x)+ C' with C'= C+ 3ln(\sqrt{3}).
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    Thats an indefinite integral so your answer should be x\sqrt{x^2+ 3}+ 3ln(\frac{\sqrt{x^2+ 3}+ x}{\sqrt{3}})+ C which is the same as x\sqrt{x^2+ 3}+ 3ln(\sqrt{x^2+ 3}+ x)- 3ln(\sqrt{3})+ C which is the same as x\sqrt{x^2+ 3}+ 3ln(\sqrt{x^2+ 3}+ x)+ C' with C'= C+ 3ln(\sqrt{3}).
    Right you are sir. Thank you
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