How to integrate 1/(x*sqrt(x-1))

Added: 2026-05-05

[00:00:00]So I sent this integral as one of my homework questions for my calculus 2 class and I thought it's pretty nice. So let's talk about it. We have the integral of 1 /x * the square<unk> of x -1.

[00:00:12]So first let me just talk about it. If we had the integral of 1 /x * the square root of x^2. If we have the square right here, this right here is easier in the sense that there's a much clearer indication of let's just do tricks up because we have the x square minus one inside the square root. Let x equal secant theta.

[00:00:36]That way we can see dx equ= secant theta tangent theta d theta and if you plug that in you will see that secant^ square theta - 1 is equal to tangent square theta. This is just a trick identity and you'll get the integral 1 / dx is secant theta time the square root. If you put this inside here, you just get tangent square theta and uh dx is all that. So seeant theta tangent theta d theta. In fact this and that cancel and you see secant theta and that cancel tangent theta tangent theta cancel. You're just integrating one in the theta world. So you just get theta. If you look back here, you see that theta is equal to inverse secant and then we're done. And in fact, if you remember the derivative of inverse secant is equal to this, then you can actually just put on the result as well.

[00:01:44]So that's okay. But now the problem is that we don't have the square right here. So what do we do? H maybe we do u sub instead. And usually if we do u equal the inside which is x minus one then we see du is equal to just dx and that will be the integral 1 / dx. Well it's not allowed in the u world but it's okay. We can add one to both sides. U equals I want to say u is equal to plus no u + 1 is equal to x. So I can put that right here. U + one parenthesis that and then square root of the inside is the U and uh we have the D U u on the side and now what you can distribute a little bit you can still finish it still doable you try it but it might not be too clear from here in my opinion the trick is you have to look at this u as square root of u squared and then do another U sub.

[00:02:52]Yeah, you can go from there. It's doable. But I want to show you though sometimes maybe you want to let u equal the whole square root. I think it works out very well. So let's have a look. All right. So for this one, I'm actually going to let u equal<unk> of x - 1. And then because we have to get the x right here and also the dx. So let's isolate the x first. Square both sides.

[00:03:19]and then add one. So x equ= u ^2 + 1.

[00:03:24]Differentiate now dx equals the derivative of this is 2 u and then d u.

[00:03:31]Now take this to the u world we get the integral dx is u ^2 + 1 and then put parenthesis around it. And then we have square root of u. Well the whole thing is u now right square root of x - 1. It's a u. So multiply by u one on the top dx is 2 u du.

[00:03:56]Now this time you see that the u and u cancel right away. Very nice. You can put a two all the way in the front and then look at the integral of 1 / u ^2 + 1.

[00:04:07]Now do we recognize the what function will give us this? Yes. Inverse tangent.

[00:04:13]Yes. So the answer is 2 * the inverse tangent of u. And I'm not going to put on plus c yet because I haven't gone back to the x world. So the last step u is the whole thing right square root of x - one and then plus c and then we are done.

[00:04:33]Very nice. In fact when you have an integral with this structure you can let u equal to the square root and it works out very well. depending on how you are going to change exponent here. So I can give you guys this one for you to try as well integral 1 /x * the square root. In fact, if you have x to a second power, letting you sub equal the square root part right here, it will also work very nicely as well. But why don't you guys try the third power because for this one the regular usab is going to beh I don't know I a lot of us will just get stuck after the equ because the expression is not going to be so nice and uh yeah the key is this whole thing is a u. So keep in mind sometimes when you're dealing with an expression maybe you have to let u equal the whole square root part maybe works out really well. Unfortunately, I have to tell you there's no one method that will solve all the integrations.

[00:05:36]That's just not possible. So, you'll just have to give these kind of things a try and hope for the best.