 In this video, we provide the solution to question number 11 for practice exam number two for math 1220, in which case we have to evaluate the indefinite integral of the function of the square root of 1 plus x over 1 minus x with respect to x here. Now on this question, there are perhaps multiple ways you could approach it. There's not necessarily one single way to do it. Some of the thoughts that come to mind when I look at this function is you do have this square root with a rational function inside of it, 1 plus x over 1 minus x. Square roots do cause a problem. Radicals in general causes a problem and many anti-derivative problems. And so there's a couple strategies one could utilize to try to help alleviate the burden of square root. One possible attempt is you do a rationalizing substitution for which case you would set your u equal to the square root of 1 plus x over 1 minus x. To do so, though, then you'd have to also take the derivative, which the derivative of a square root is going to be 1 over 2 times that very square root. But then you also have to take the inner derivative for which you're going to take the derivative of 1 plus x over 1 minus x. That's going to use the quotient rule DX. And so while you can use the substitution to remove this square root in terms of u, however messy this turns out might be difficult to make the rationalizing substitution work. Because the rationalizing substitution is really only going to work if I can address all of the x's in terms of u. So while there is some promise here, there is a little bit of hope. It doesn't seem like a really great hope. I'm afraid that the derivative of this thing is going to complicate it dramatically. So I'm going to actually plan a different strategy here. The second strategy that we have for dealing with square roots typically is with regard to a trigonometric substitution. Now trig subs are great when you have like a sum or difference of squares inside of the square root. But I don't have that here. I have a 1 plus x on top. I have a 1 minus x on the bottom. But it turns out what we can do is if I can actually multiply top and bottom and you can pick. You can multiply top and bottom by 1 plus x or 1 minus x. We can actually produce some squares here in the following manner. So let's see what's exactly in play here. I'm going to take this 1 plus x and this 1 minus x. And then inside of the square root, I'm going to times top and bottom by 1 plus x. So this is intended to be inside the square root still. Notice that by timesing by 1 plus x over 1 plus x, this fraction is just the number 1. It's a strategic number 1 and that's going to help us out here because in the numerator, notice what happens. In the numerator, you're going to get 1 plus x times 1 plus x. I'm not going to foil it out. I'm going to leave that as a 1 plus x squared because this is inside of a square root. That's going to cancel out now in the denominator. This is what was nice here because notice I have a 1 plus x and a 1 minus x. Like they're the same expression, but the sign is different in the denominator. I'm going to multiply that out and then I'm going to get the square root of 1 minus x squared dx. So this then simplifies to be the integral of 1 plus x over the square root of 1 minus x squared dx. So I've algebraically rewritten the function in such a format that I think a trig sub is going to be much more advantageous than it was before. Looking at the square root there, we have the square root of 1 minus x squared. That indicates to me that x should equal sine of theta. Therefore, dx is equal to cosine of theta and the square root of 1 minus x squared is equal to likewise cosine theta. I did forget my d theta right there. Don't forget that. And so then I can rewrite my function. I'll actually write this in green now that I'm doing a trig sub. You're going to get 1 plus x, which x is sine. This sits on top of the square root of 1 minus x squared, which then becomes a cosine. And then your dx, excuse me, is a cosine theta d theta. Notice in this situation, you have a cosine theta here cancels with a cosine theta. And so this will simplify just to be the integral of 1 plus sine theta d theta, which is pretty nice. The antiderivative there will be theta plus. I need the antiderivative of sine, which is actually a negative cosine. So you get a negative cosine theta plus a constant. Don't forget the constant there. So now we have to turn the theta's back into x's. Notice that cosine theta is equal to the square root. So that's a nice substitution there. In order to get theta by itself, take the original substitution and solve for theta that's going to involve an arc sine. So our antiderivative is going to be arc sine of x. Minus the square root of 1 minus x squared plus C. And you can see here that because the antiderivative involves an arc sine, basically the only method that's going to be super helpful here is going to be a trig substitution. Anything else is ultimately going to lead to a trig sub probably. I can see that now with the antiderivative. So that rationalizing substitution probably was not going to work if I had tried to continue that onward here. And just also just one comment about the original one. I noticed how I did this 1 plus x times 1 plus 1 plus x on top and bottom. I actually preferred the square root to be in the denominator so that there's some more cancellation that's better here. But you could have done it the other way around. You could have times 1 minus x on top, 1 minus x on the bottom. If you had done that, you would have ended up with a square root of 1 minus x squared over 1 minus x. Turns out this would be fine, but it's a little bit cleaner, a little bit easier if you put the square root in the denominator as opposed to the difference 1 minus x in the denominator. The square root is better in the denominator. That's why I chose the path that I did here.