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Math Help - Integration Problem.

  1. #1
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    Integration Problem.

    Find the integral of
    {{(x4 + 1)1/2 }[ ln( x2 + 1 ) + 2lnx]} /x​4
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  2. #2
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    Re: Integration Problem.

    Quote Originally Posted by sohamjariwala View Post
    Find the integral of
    {{(x4 + 1)1/2 }[ ln( x2 + 1 ) + 2lnx]} /x​4
    It doesn't have a closed form solution. Maybe try a series solution instead...
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  3. #3
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    Re: Integration Problem.

    I think I made a mistake in writing it

    the correct question is
    {{x2 + 1)1/2[ ln(x2 + 1) - 2lnx]}/x​4
    Last edited by sohamjariwala; July 12th 2012 at 09:41 PM.
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  4. #4
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  5. #5
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    Re: Integration Problem.

     \text{Integrate:}\mspace{7mu} \int{\frac{\sqrt{x^2+1}}{x^4}}\left(\ln{(x^2+1)}-2\ln{x}\right)dx}\quad \quad (1)

     (1)\, \Leftrightarrow\, \int{\frac{\sqrt{x^2+1}}{x^4}}\ln\left({\frac{x^2+  1}{x^2}}\right)dx}\, \Leftrightarrow\, \int{\frac{1}{x^3}\sqrt{\frac{x^2+1}{x^2}}\ln\left  ({\frac{x^2+1}{x^2}}\right)dx}

     \text{Let}\mspace{6mu} u=\sqrt{\frac{x^2+1}{x^2}}\, \Leftrightarrow\, u^2=\frac{x^2+1}{x^2}=1+\frac{1}{x^2}>0

     \Rightarrow\, 2udu=-\frac{2}{x^3}dx\, \Leftrightarrow\, dx=-x^3udu

     \int{\frac{1}{x^3}\cdot u\cdot \ln{u^2}\cdot (-x^3udu)}\, \Leftrightarrow\, -2\int{u^2\ln{(u)}du}\quad \quad (2)

     \text{Integration by parts}

     \begin{array}{lllllll} w=\ln{u} &&& & & dv=u^2du\\ dw=\dfrac{1}{u} &&& & & v=\dfrac{u^3}{3}\\ \end{array}

     (2)\, \Leftrightarrow\, -\frac{2}{3}\left( u^3\ln{u}-\int{u^2}du \right) = -\frac{2}{9}u^3 \left( 3\ln{u}-1 \right) = -\frac{2}{9}u^3 \left( \ln{\frac{u^3}{e}} \right)

     \text{No need to have the absolute value sign for \emph{u} in the log term since}\,u>0.

     \Leftrightarrow\, -\frac{2}{9} \left( \frac{x^2+1}{x^2}\right)^{3/2} \ln{\left[ \frac{1}{e}\left(\frac{x^2+1}{x^2}\right)^{3/2} \right]}

     \Leftrightarrow\, -\frac{2}{9} \cdot \frac{\left(x^2+1\right)^{3/2}}{x^3} \ln{\frac{\left(x^2+1\right)^{3/2}}{ex^3}}\, \Leftrightarrow\, \frac{2}{9} \cdot \frac{\left(x^2+1\right)^{3/2}}{x^3} \ln{\frac{ex^3}{\left(x^2+1\right)^{3/2}}}

     \int{\frac{\sqrt{x^2+1}}{x^4}}\left(\ln{(x^2+1)}-2\ln{x}\right)dx} = \frac{2}{9} \cdot \frac{\left(x^2+1\right)^{3/2}}{x^3} \ln{\frac{ex^3}{\left(x^2+1\right)^{3/2}}} + C

     \text{Time for some tea!}
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  6. #6
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    Re: Integration Problem.

    Superb,My friend thank you.
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  7. #7
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    Re: Integration Problem.

    I also have another problem. Can you figure out what to substitute.
    Integrate w.r.t. x
    1/[(cot(x/2)cot(x/3)cot(x/6)]
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