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Math Help - Easy Integrals!!

  1. #1
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    Easy Integrals!!

    Evaluate the following integrals

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  2. #2
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    1st one comes out to:
     \frac{1}{4}x^2 (2log(x) - 1)

    2nd one comes out to:
     \frac{1}{5}(2log(x - 2) + 3log(x + 3))

    I think! not 100% certain there.
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by matty888 View Post
    Evaluate the following integrals


    for the first one use the change of base formula to get

    \int_{1}^{2}x\log(x)dx=\int_{1}^{2}x\frac{\ln(x)}{  \ln(10)}dx=\frac{1}{\ln(10)}\int_{1}^{2}x\ln(x)dx

    use integration by parts with u=ln(x) dv=x

    That should get you started
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by matty888 View Post
    Let's find the partial fraction decomposition of \frac{x}{x^2+x-6}.

    The discriminant of the denominator is  \Delta = 1^2-4\cdot1\cdot (-6)=25>0 hence the denominator has to different real roots : \frac{-1\pm\sqrt{25}}{2} that is to say -3 and 2. Hence x^2+x-6=(x+3)(x-2)

    It gives <br />
\frac{x}{x^2+x-6} = \frac{x}{(x+3)(x-2)}=\frac{\alpha}{x+3}+\frac{\beta}{x-2} which has to be solved for \alpha and \beta.

    Can you take it from here ?
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  5. #5
    Math Engineering Student
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    \begin{aligned}<br />
   \frac{x}{x^{2}+x-6}&=\frac{2x+1-1}{2(x-2)(x+3)} \\ <br />
 & =\frac{1}{2}\left[ \frac{2x+1}{x^{2}+x-6}-\frac{(x+3)-(x-2)}{5(x-2)(x+3)} \right] \\ <br />
 & =\frac{1}{2}\left[ \frac{2x+1}{x^{2}+x-6}-\frac{1}{5}\left\{ \frac{1}{x-2}-\frac{1}{x+3} \right\} \right].<br />
\end{aligned}

    Integrate.

    P.S.: first integral can be tackled by double integration technique too.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    \begin{aligned}<br />
   \frac{x}{x^{2}+x-6}&=\frac{2x+1-1}{2(x-2)(x+3)} \\ <br />
 & =\frac{1}{2}\left[ \frac{2x+1}{x^{2}+x-6}-\frac{(x+3)-(x-2)}{5(x-2)(x+3)} \right] \\ <br />
 & =\frac{1}{2}\left[ \frac{2x+1}{x^{2}+x-6}-\frac{1}{5}\left\{ \frac{1}{x-2}-\frac{1}{x+3} \right\} \right].<br />
\end{aligned}

    Integrate.
    Haha! i LOVE seeing when you do that!
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