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Thread: 3 really hard integrals

  1. #1
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    3 really hard integrals

    Hi! I would really appreciate any help solving couple of integrals. I probably wasted weeks on them... and small forest worth of paper...

    1st should be solved with u-substitution;
    \int \frac{2^x\cdot 3^x}{9^x+4^x}dx

    \int \frac{x^3-6}{x^4+6x+8}dx

    \int \frac{\sqrt{2x-1}+\sqrt{4x-1}+1}{\sqrt{2x-1}-\sqrt{4x+1}}dx

    If anyone have any idea how to solve them, because they all appear unsolvable to me (or anyone I asked for help so far).
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  2. #2
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    Re: 3 really hard integrals

    Some suggestions:
    1) Factor $9^x=(3^x)^2$ out of the numerator and denominator. You then have something of the form $1+f^2(x)$ in the denominator which may respond to a substitution of $\tan u = f(x)$.

    3) Multiply top and bottom by the conjugate of the denominator $\sqrt{2x-1} + \sqrt{4x+1}$
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  3. #3
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    Re: 3 really hard integrals

    1) still working on it (and getting nowhere near solution)
    3) so after multiplication of denominator I get \int \frac{2x-1+(\sqrt{4x-1}+1)(\sqrt{2x-1}+\sqrt{4x+1})+\sqrt{2x-1}-\sqrt{4x+1}}{-2x-2}}dx is this correct and if it is, then what next?
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  4. #4
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    Re: 3 really hard integrals

    Your denominator is wrong (as the very least), it should be a single term.
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    Re: 3 really hard integrals

    $$\newcommand{integral}[2][x]{\int {#2} \, \mathrm d #1}\begin{aligned}
    &&\integral{2^x 3^x \over 9^x + 4^x} &= \integral{{1 \over 3^{2x}} \cdot {2^x 3^x \over 1 + \left({2^{2x} \over 3^{2x}}\right)}} \\
    &&&= \integral{ {{2^x \over 3^x} \over 1 + \left({2 \over 3}\right)^{2x}}} \\
    &&&= \integral{\big({2 \over 3}\big)^x \over 1 + \bigg(\big({2 \over 3}\big)^x\bigg)^2} \\
    &&&= \integral{a^x \over 1 + \left(a^x\right)^2} &\text{where $a = \tfrac23$} \\
    &\text{Let $u = a^x$ so that $\mathrm d u = a^x \log a \, \mathrm d x$ and}& \integral{2^x 3^x \over 9^x + 4^x} &= {1 \over \log a}\integral[u]{1 \over 1 + u^2}
    \end{aligned}$$
    Can you fill in the gaps and complete the integral?
    Last edited by Archie; Aug 21st 2015 at 08:19 AM.
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  6. #6
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    Re: 3 really hard integrals

    For 2) you may notice that you have $$\int {\tfrac14 f'(x) + c \over f(x)} \mathrm d x$$
    This can be split into two integrals, the first of which is a standard result. The second requires the factoring of the denominator and then solve using the method of partial fractions.

    The factorisation is going to be of the form $x^4 + 6x + 8 = (x^2 + ax + b)(x^2 + cx + d)$. You can expand the right hand side of that equation and compare coefficients (of $x^k$ for $0 \le k \lt 4$) to get four equations in four unknowns.
    Last edited by Archie; Aug 21st 2015 at 08:16 AM.
    Thanks from confuseddove and sakonpure6
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    Re: 3 really hard integrals

    In the end I got \frac{1}{ln(\frac{2}{3})}arctan(\frac{2}{3})^x)+c.
    I'll try from scratch in other 2...

    Thank you Archie
    Last edited by confuseddove; Aug 21st 2015 at 09:14 AM.
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    Re: 3 really hard integrals

    I am still fighting battles with these 2 integrals.
    I really tried (but failed) to find a b c d for factorisation for 2),... I somehow managed to get to the end with integration with parts.
    And in 3) I got some spaghetti monster, over 5 pages long.
    But my solution for both integrals is wrong(took derivative and it is not same class of functions), some silly mistake probably... I would really appreciate anyone showing me solution for this.
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    Re: 3 really hard integrals

    Is the second supposed to be $x^4 + 6x^2 + 8 = (x^2 + 4)(x^2 + 2)$ in the denominator?

    The third one seems likely to have an error in too. Here's the Wolfram Alpha solution.
    Last edited by Archie; Aug 24th 2015 at 07:06 PM.
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    Re: 3 really hard integrals

    It might be typo in textbook and that would certainly explain why both of them are such a pain in the ***. I'll ask professors at uni when I see him...

    I tried with Mathematica from rpi, maple, symbolab and wolfram alpha and none make any real sense from both.

    Thank you for all your assistance I really appreciate it.

    If anyone finds a challenge in this, post a solution here, but please don't loose your sleep over this for me, it is not worth it!
    Last edited by confuseddove; Aug 24th 2015 at 10:54 PM.
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  11. #11
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    Re: 3 really hard integrals

    Quote Originally Posted by confuseddove View Post
    It might be typo in textbook and that would certainly explain why both of them are such a pain in the ***. I'll ask professors at uni when I see him...

    I tried with Mathematica from rpi, maple, symbolab and wolfram alpha and none make any real sense from both.

    Thank you for all your assistance I really appreciate it.

    If anyone finds a challenge in this, post a solution here, but please don't loose your sleep over this for me, it is not worth it!
    What was the textbook? I have a professional interest in the location of possible mistakes in mathematics textbooks. I may be able to research it -- on the other hand I can't promise.
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  12. #12
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    Re: 3 really hard integrals

    Well actually it is not textbook, just pdf with example problems. It is intended for students to prepare for written part of exam.
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