How Do You Integrate (x + 3) / sqrt(x^2 + 4x - 5)?

[renaldocoetz]

Homework Statement

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

Homework Equations

The Attempt at a Solution

No idea. they are not derivative / integral of each other so I am not sure :( Please help asap. Thanks in advance

 

[Hootenanny]

Homework Statement

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

Homework Equations

The Attempt at a Solution

No idea. they are not derivative / integral of each other so I am not sure :( Please help asap. Thanks in advance

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.

 

[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 >.<

 

[Hootenanny]

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

For the very reason you say:

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?

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?

 

[renaldocoetz]

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?

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?

 

[Hootenanny]

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?

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.

 

[HallsofIvy]

There is a single obvious substitution for both integrals.

 

[Hootenanny]

There is a single obvious substitution for both integrals.

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.

 

[renaldocoetz]

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

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] ?

 

[Hootenanny]

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

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.

 

[vela]

I'm not sure what standard integral you are referring to, but that is not the correct answer.

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.

 

[Hootenanny]

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.

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.

 

[vela]

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.

 

[Hootenanny]

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.

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