# Integral from Hell

1. Feb 26, 2009

### lubuntu

Integral from Hell!!

1. The problem statement, all variables and given/known data

$$\int \frac{\sqrt{ln(9-x}}{\sqrt{ln(9-x} + \sqrt{ln(x+3)}} dx$$

2. Relevant equations
woe = me

3. The attempt at a solution
Tried all the algerbraic manipulations I could think of, can't figure out a substitution either.

2. Feb 26, 2009

### letmeknow

Re: Integral from Hell!!

Have you tried multipying by the conjugate? Just a guess ( not a hint guess, but a real guess)

3. Feb 26, 2009

### lubuntu

Re: Integral from Hell!!

Tried didn't get me anywhere seemingly

4. Feb 26, 2009

### gabbagabbahey

Re: Integral from Hell!!

Is this supposed to be an indefinite integral, or are you given limits? Also, are you sure it is exactly as written above?

5. Feb 26, 2009

### lubuntu

Re: Integral from Hell!!

limits from 2 to 4, figured i'd just get help getting the anti derivative

6. Feb 26, 2009

### gabbagabbahey

Re: Integral from Hell!!

I don't think there is a general antiderivative; but I suspect that the limits allow you to use some trick (like in the 1/(1+tan^b (x) ) thread).....hmmmm, let me see....

7. Feb 26, 2009

### lubuntu

Re: Integral from Hell!!

Holy crap this is linear! anyone care to help me figure out why?

8. Feb 26, 2009

### gabbagabbahey

Re: Integral from Hell!!

Really?!...Doesn't look exactly linear to me....It looks close to a straight line, but not really.

9. Feb 26, 2009

### lubuntu

Re: Integral from Hell!!

I could have been mistaken

10. Feb 26, 2009

### gabbagabbahey

Re: Integral from Hell!!

I'd start by dividing the numerator and denominator by $\sqrt{\ln(9-x)}$...

11. Feb 26, 2009

### lubuntu

Re: Integral from Hell!!

nice one! how didn't i see that!!

12. Feb 26, 2009

### gabbagabbahey

Re: Integral from Hell!!

Lol , actually, I'm still trying to see why the area under $$\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$ is the same as the area under $$\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}$$

Is there some proof I'm not remembering?

13. Feb 26, 2009

### lubuntu

Re: Integral from Hell!!

not that i know of, i'm kind of stuck at that point now I don't see any substitutions that would make things simpler from here.

14. Feb 26, 2009

### lubuntu

Re: Integral from Hell!!

$$\frac{\frac{\ln(3+x)}{\ln(9-x)}^b}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$

15. Feb 26, 2009

### lubuntu

Re: Integral from Hell!!

I dont see that going anywhere, anyone got ideas?

16. Feb 26, 2009

### gabbagabbahey

Re: Integral from Hell!!

Sure, $$\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}= \frac{\frac{\ln(3+x)}{\ln(9-x)}^b}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$

so $$\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}+\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}=1$$

But that doesn't explain why the area under $$\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$ is equal to the area under $$\frac{1}{1+\left(\frac{\ln(9-x)}{\ln(3+x)}\right)^b}$$ over that interval (although I know from mathematica that it is)

17. Feb 26, 2009

### gabbagabbahey

Re: Integral from Hell!!

Hmmm....what do you get when you flip the function $$\frac{1}{1+\left(\frac{\ln(3+x)}{\ln(9-x)}\right)^b}$$ over the line $x=3$?

You may need to look in your linear algebra text, under the transformations/reflections section for this one)

18. Feb 26, 2009

### lubuntu

Re: Integral from Hell!!

I think I'm giving up!! This was one of 4 challenge problems my professor gave... I think I got 2 of the others I don't know on this one haha.

19. Feb 26, 2009

### gabbagabbahey

Last edited by a moderator: May 4, 2017
20. Feb 26, 2009

### lubuntu

Re: Integral from Hell!!

Of course not, nice job finding that out, not sure if it is gonna help me compute the integral.