# Homework Help: Calculating indefinite integral

1. Dec 12, 2011

### Argiris

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

hey could you help me to calculate the indefinite integral of y=√(x+1)/√(x+2)

2. Relevant equations

3. The attempt at a solution
tried to set x+1=u and integrate it by substitution but didnt work
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Dec 12, 2011
2. Dec 12, 2011

### micromass

Try to set $u^2=x+1$

3. Dec 12, 2011

### Ray Vickson

It is not clear what your function y is. It could be $$y = \sqrt{x} + \frac{1}{x} + 2,$$ or $$y = \sqrt{\displaystyle x + \frac{1}{x}} + 2,$$ or
$$y = \sqrt{\displaystyle x + \frac{1}{x} + 2}.$$ If you don't want to use LaTeX, you need to use brackets, so the first way I wrote above would be y = (√x) + (1/x) + 2, the second way would be y = √[x + (1/x)] + 2, and the thire way would be y = √[x + (1/x) + 2]. If I read your function using *standard* rules and priorities, it means the first way above.

RGV

4. Dec 12, 2011

### daveb

So now you have ∫√u/(u+1)du. Try an additional substitution.

5. Dec 12, 2011

### Argiris

I apologize for the missunderstanding the function is y=Sqrt[x+1]/Sqrt[x+2] i corrected it in the question to.

6. Dec 12, 2011

### Argiris

well i tried it and this transformed the integral into ∫2*(u^3)/√(u^2+1)du. then i set u=tanθ and the integral is transformed into 2∫(tanθ^3)*secθdθ. and couldnt take it any further..

7. Dec 12, 2011

### dextercioby

$u = \sinh t$ seems a better substitution.

8. Dec 13, 2011

### Staff: Mentor

EDIT: double posting. see next post.

Last edited: Dec 13, 2011
9. Dec 13, 2011

### Staff: Mentor

I don't get that u^3.

10. Dec 13, 2011

### clanijos

Indeed sinh does seem like a better option here

11. Dec 13, 2011

### SammyS

Staff Emeritus
Similar, but slightly different than what micromass suggested.

Use the substitution: $u=\sqrt{x+2}\,,$ then $\displaystyle du=\frac{dx}{2\sqrt{x+2}}\,.$

This also gives $\sqrt{x+1}=\sqrt{u^2-1}\,.$

12. Dec 14, 2011