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Math Help - Find the indefinite integral

  1. #1
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    Find the indefinite integral

    Hi,

    \displaystyle\int\frac{4}{4x^2+4x+65}dx

    Hints please. I am stumped by this problem. Maybe completing the square?
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  2. #2
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    Yes, complete the square...

    \displaystyle \int{\frac{4}{4x^2 + 4x + 65}\,dx} = \int{\frac{1}{x^2 + x + \frac{65}{4}}\,dx}

    \displaystyle = \int{\frac{1}{x^2 + x + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2 + \frac{65}{4}}\,dx}

    \displaystyle = \int{\frac{1}{\left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{65}{4}}\,dx}

    \displaystyle = \int{\frac{1}{\left(x + \frac{1}{2}\right)^2 + 16}\,dx}.


    Now make the substitution \displaystyle x + \frac{1}{2} = 4\sinh{t} so that \displaystyle dx = 4\cosh{t}\,dt and the integral becomes

    \displaystyle \int{\frac{1}{(4\sinh{t})^2 + 16}\,4\cosh{t}\,dt}

    \displaystyle = \int{\frac{4\cosh{t}}{16(\sinh^2{t} + 1)}\,dt}

    \displaystyle = \int{\frac{\cosh{t}}{4\cosh{t}}\,dt}

    \displaystyle = \int{\frac{1}{4}\,dt}

    \displaystyle = \frac{t}{4} + C

    \displaystyle = \frac{\sinh^{-1}\left(\frac{x}{4} + \frac{1}{8}\right)}{4} + C.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by ugkwan View Post
    Maybe completing the square?
    Yes,

    4x^4+4x+65=4(x+\;1/2)^2+64

    Now use t=2(x+\;1/2) , etc.

    Regards.

    Sorry Prove It, I didn't see your answer.
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  4. #4
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    Or: from \displaystyle \int{\frac{1}{\left(x + \frac{1}{2}\right)^2 + 16}\,dx}
    let 4u= x+ 1/2, so that 4du= dx and the integral becomes
    \displaystyle \int{\frac{1}{\left(16u^2 + 16}\, 4du}= \frac{1}{4}\int\frac{1}{u^+ 1}du
    \displaystyle = \frac{1}{4}arctan(u)+ C
    \displaystyle= \frac{1}{4}arctan\left(\frac{1}{4}(x+ 1/2)\right)+ C
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Or: from \displaystyle{ \int{\frac{1}{\left(x + \frac{1}{2}\right)^2 + 16}\,dx}
    let 4u= x+ 1/2, so that 4du= dx and the integral becomes
    \displaystyle \int{\frac{1}{\left(16u^2 + 16}\, 4du}= \frac{1}{4}\int\frac{1}{u^+ 1}du
    \displaystyle = \frac{1}{4}arctan(u)+ C
    .
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