1. Aug 29, 2011

### renaldocoetz

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

Integrate: (x + 3) / sq rt of (x2 + 4x - 5)

2. Relevant equations

3. The attempt at a solution

Last edited: Aug 29, 2011
2. Aug 29, 2011

### Hootenanny

Staff Emeritus
The first step is to complete the square on the argument of the square root, and then look for for an appropriate substitution.

P.S. Please do not post duplicate threads, it only serves to clutter up the forums.

3. Aug 29, 2011

### renaldocoetz

that gives me a denominator of (x+2)2 - 9 ... why do we do this?

also: can I split the initial fraction up into 2 separate ones? IE. (x+2) and (+1) (both over same denominator) in this way I can use substitution for the 1st fraction right?

PS. sorry about the duplicate post. Realized it was in the wrong section >.<

4. Aug 29, 2011

### Hootenanny

Staff Emeritus
For the very reason you say:
So, you will have

$$I = \int \frac{x+2}{\sqrt{(x+2)^2 - 9}}\;\text{d}x +\int \frac{1}{\sqrt{(x+2)^2 - 9}}\;\text{d}x$$

Now, what do you think would be a good substitution?

5. Aug 29, 2011

### renaldocoetz

well before completing the square it seemed quite clear to me that letting u= x2+ 4x -5 would make 1/2 du = x+2. for solving the 1st fraction.. but looking at the completed square i assume i should be doing something else?

6. Aug 29, 2011

### Hootenanny

Staff Emeritus
Of course you could have split the integrand into partial fractions with completing the square and then used your substitution on the first fraction. However, you would need to complete the square for the second fraction anyway. For the first integral, you are already in a "canonical form", i.e. you have

$$I_1 = 2\int \frac{du}{u}$$

For the second, try letting $3\sec\theta = (x+2)$ and see where that takes you.

Last edited: Aug 29, 2011
7. Aug 29, 2011

### HallsofIvy

Staff Emeritus
There is a single obvious substitution for both integrals.

8. Aug 29, 2011

### Hootenanny

Staff Emeritus
I would say the first could be considered obvious (of course it is only obvious if one spots it). However, I would contest that the second integral is by no means obvious.

9. Aug 29, 2011

### renaldocoetz

took me to a world of confusion :( sheesh what am I missing?

also, can one use the standard integral with the second fraction and get arc cosh [(x+2)/3] ?

10. Aug 29, 2011

### Hootenanny

Staff Emeritus
I'm not sure what standard integral you are referring to, but that is not the correct answer. I assume that you are okay with the first integral. So, for the second, let $3\sec\theta = (x+2)$ such that $dx = 3\sec\theta\tan\theta d\theta$. Thus, the integral becomes

$$I_2 = 3\int{\frac{\sec\theta\tan\theta}{3\sqrt{\sec^2 \theta -1}}}\text{d}x$$

If you then recall that $\tan^2\theta = \sec^2\theta -1$, the integral should now be straightforward.

11. Aug 29, 2011

### vela

Staff Emeritus
I think the OP is referring to using an integral table. The arccosh answer is in fact correct. You can use the substitution $x+2 = 3 \cosh u$ to derive it.

12. Aug 29, 2011

### Hootenanny

Staff Emeritus
Perhaps I am missing something because, I have (following from my previous post)

$$I_2 = \int \sec\theta \text{d}\theta$$

which yields,

$$I_2 = \log(\tan\theta + \sec\theta)$$

And eventually

$$I_2 = \log([x+2]+\sqrt{(x+2)^2 - 9})\;.$$

Now, the proposed solution $\text{arccosh}[(x+2)/3]$ has a root at x=1. However, clearly $I_2$ is non-zero at x=1.

Last edited: Aug 29, 2011
13. Aug 29, 2011

### vela

Staff Emeritus
The two solutions differ by a constant. If you express arccosh in terms of log, you'll find you can convert one solution into the other.

14. Aug 29, 2011

### Hootenanny

Staff Emeritus
Well, I'll be damned. I entirely missed that one!