# Thread: 3 really hard integrals

1. ## 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).

2. ## 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 $\sqrt{2x-1} + \sqrt{4x+1}$

3. ## 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?

4. ## Re: 3 really hard integrals

Your denominator is wrong (as the very least), it should be a single term.

5. ## 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?

6. ## 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.

7. ## 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

8. ## 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.

9. ## 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.

10. ## 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!

11. ## Re: 3 really hard integrals

Originally Posted by confuseddove
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.

12. ## 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.