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Thread: Hard Integration help

  1. #1
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    Hard Integration help

    Can you please solve this problems?

    $\displaystyle \int\frac{x^4-x {dx}}{x^2-1}$
    $\displaystyle \int\frac{e^{4y}}{e^{2y}+1}dy$
    $\displaystyle \int\frac{dx}{(2x-1)[3-2 \ln (1-2x)]}$
    $\displaystyle \int\frac{dx}{x(1-x^3)}$
    $\displaystyle \int\frac{(3-4x)x dx}{\sqrt[3]{1-x}}$


    Thanks in Advance
    Last edited by mr fantastic; Aug 3rd 2009 at 04:25 AM. Reason: Fixed exponents in second integral
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  2. #2
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    Quote Originally Posted by Yuroichi View Post
    Can you please solve this problems?

    $\displaystyle \int\frac{x^4-x {dx}}{x^2-1}$
    $\displaystyle \int\frac{e^{4y}}{e^{2y}+1}dy$
    $\displaystyle \int\frac{dx}{(2x-1)[3-2 \ln (1-2x)]}$
    $\displaystyle \int\frac{dx}{x(1-x^3)}$
    $\displaystyle \int\frac{(3-4x)x dx}{\sqrt[3]{1-x}}$


    Thanks in Advance
    1. Note that $\displaystyle \frac{x^4-x}{x^2-1} = \frac{x(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} = \frac{x^3 + x^2 + x}{x + 1} = x^2 + 1 - \frac{1}{x + 1}$.

    2. Start by substituting $\displaystyle u = e^{2y}$.

    3. Start by substituting $\displaystyle u = \ln (1 - 2x)$.

    4. Completely factorise the denominator and then use a partial fraction decomposition.

    5. Start by substituting $\displaystyle u^3 = 1 - x$.

    If you need more help, please post your working and state where you get stuck.
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