# Tricky Antiderivative

1. Mar 19, 2013

### pierce15

1. The problem, the whole problem, and nothing but the problem

$$\int \sqrt{\frac{1+x^2}{1-x^2}} dx$$

2. Relevant equations

?

3. The attempt at a solution

I tried a partial fraction decomp, but this obviously didn't get me very far because this isn't a rational function. Is there some kind of ingenious substitution applicable here?

2. Mar 19, 2013

### Dick

I don't see any way to get an antiderivative. And Wolfram Alpha says it's an elliptic integral. I don't disagree. If it's a definite integral there might be a trick. Is that the whole problem?

3. Mar 19, 2013

### pierce15

I've changed the problem to the following:

$$\int \sqrt{\frac{1+x}{1-x}} dx$$

According to wolfram, this does have a (nasty) elementary antiderivative. Any thoughts?

4. Mar 19, 2013

### Dick

You are making up HARD PROBLEMS.

5. Mar 19, 2013

### pierce15

Sorry for adding new detail to a previous post by the way, I just remembered that that's a big no-no

6. Mar 19, 2013

### iRaid

Making your own integrals isn't a good idea because you tend not to know if you can solve them. For example: $\int e^{x^{2}} dx$

7. Mar 20, 2013

### runningninja

The new integral you made up is solvable however. Do a u substitution.

8. Mar 20, 2013

### Curious3141

Substitute something that would make that square root on the denominator much easier to deal with (or eliminate).

9. Mar 20, 2013

### pierce15

How about the substitution $x = -2u - u^2$, $dx = -2 - 2u$

$$\int \sqrt{ \frac{1 - 2u-u^2}{1+2u + u^2} } (-2-2u) \, du$$

$$-2 \int \frac{ \sqrt{1-2u-u^2} }{1+u}(1+u) \, du$$

$$-2 \int \sqrt{1-2u-u^2} \, du$$

Alright, that looks fairly simple, but I'm blanking as to how I should deal with it.

Last edited: Mar 20, 2013
10. Mar 20, 2013

### Dick

I'd say complete the square and then use a trig substitution.

11. Mar 20, 2013

### Curious3141

Why not $x = 1 - u^2$?

12. Mar 20, 2013

### pierce15

Alright, I completed the square, changing $1-2u-u^2$ to $2-(1+u)^2$.

$$-2\int \sqrt{2-(1+u)^2} \, du$$

$$1+u = \sqrt{2} \sin\theta, \quad du = \sqrt{2} \cos \theta \, d\theta$$

$$-2\sqrt{2} \int \cos \theta \sqrt{2-2 \sin^2\theta} \, d\theta$$

$$-4 \int \cos \theta \sqrt{1-\sin^2\theta} \, d\theta$$

$$v = \sin \theta, \quad dv = \cos \theta \, d\theta$$

$$-4 \int 1 - v^2 \, dv$$

$$-4( v - v^3/3)$$

$$4(v^3/3 - v)$$

Before I go back and perform the various substitutions, is that correct?

13. Mar 20, 2013

### pierce15

That would have truly been a wise choice

14. Mar 20, 2013

### runningninja

Isn't 1 - sin2θ just cos2θ?

15. Mar 20, 2013

### pierce15

-.- yes

$$-4 \int \cos \theta \sqrt{1-\sin^2\theta} \, d\theta$$
$$-4 \int \cos^2 \theta \, d\theta$$
$$-4 \int \frac{1+\cos(2\theta)}{2} d\theta$$
$$-2\Big(\theta + \int \cos(2\theta) d\theta \Big)$$
$$-2(\theta + \frac{ \sin(2\theta)}{2})$$
$$- \sin(2\theta) -2\theta$$