1. Apr 17, 2013

### omri_mar

hi guys!
can you please help me with this integral? Ive tried some subs but i realy dont know how to start....
Thank you !!!!

#### Attached Files:

• ###### 72837_546314312086147_11364107_n.jpg
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2. Apr 17, 2013

### Staff: Mentor

Expanding with $\sqrt{\ln(9-x)}-\sqrt{\ln(3+x)}$ gives
$$\frac {\sqrt{\ln(9-x)}\left(\sqrt{\ln(9-x)}-\sqrt{\ln(3+x)}\right)} {\ln(9-x)-\ln(3+x)}$$

That can be split into two parts, where one part looks easy and the other part can be simplified significantly. I don't know if that will lead to a solution, however.

3. Apr 17, 2013

### omri_mar

first of all thank you!
Ive tried this and it lead me nowhere..

4. Apr 17, 2013

### Staff: Mentor

Hmm okay, I don't see how to solve the remaining part either.

An interesting observation: 3+x goes from 5 to 7, while 9-x goes from 7 to 5. There could be some symmetry to exploit, but I did not find it yet.

5. Apr 18, 2013

### omri_mar

Thats actually very interesting because according to the answers the answer is 1.
Can someone help me...?

6. Apr 18, 2013

### Curious3141

Hint: substitute $\displaystyle x = 6-y$.

EDIT: numerous glitches in the PF software are making it difficult to post. Basically, I wanted to add: don't try to simplify the integrand, etc. Just do the sub, and see what happens to the bounds. This is actually a very simple problem in disguise.

Last edited: Apr 18, 2013
7. Apr 18, 2013

### omri_mar

I cant remove the disguise... :(

8. Apr 18, 2013

### omri_mar

this is what i get

#### Attached Files:

• ###### integral.png
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9. Apr 18, 2013

### Staff: Mentor

Ah nice, that is the right way to solve it.
@omri_mar: You can modify this new integrand to look similar to the old one now.

10. Apr 18, 2013

### omri_mar

I dont get it. How can I continue?

11. Apr 18, 2013

### cpt_carrot

Try adding together the new integral and the old one

12. Apr 18, 2013

### omri_mar

what do you mean?

13. Apr 18, 2013

### Staff: Mentor

You have an equation (old integral)=(new integral). Try to write (new integral) as (something)-(old integral). This gives you an equation like 2*(old integral)=(something), and (something) is easy to calculate.

14. Apr 19, 2013

### Curious3141

Sorry for the late reply - I just noticed your bounds are off in the attachment to post 8. Your bounds should be reversed (identical to the original integral). Don't forget that $dy = -dx$. Other than that, the other posters have already told you how to proceed. Remember that in a definite integral, the variable of integration is basically just a dummy variable, so whether it's x or y, it doesn't matter. So just replace the y in the new integral with x, and add it to the original integral. Now you have twice the original integral, and it's equal to something very simple.

15. Apr 19, 2013

### SammyS

Staff Emeritus
(Most of post deleted in Edit. It was pretty much redundant with Curious3141 post. )

Now modify the numerator by adding and subtracting ... (something clever) .

Last edited: Apr 19, 2013