Integration using sinh^2 substitution

[vesu]

Homework Statement

\int \frac{1}{\sqrt{x + x^2}} dx
We have been told to use the substitution x = \sinh^2{t}.

Homework Equations

\int \frac{1}{\sqrt{a^2 + x^2}}dx = \sinh^{-1}(\frac{x}{a}) + C
Maybe?

The Attempt at a Solution

I'm not really sure where to start, we haven't done any questions involving letting x = \sinh^2{t}, only x = \sinh{t}. Substituting x into the integral doesn't seem to get me anywhere. I feel like I might have to derive x which gives \sinh{2t} but I don't know what to do with that.

 

[kontejnjer]

Try completing the square in the denominator and see what you end up with.

 

[vesu]

Ah, I think I've got it. I ended up with 2\sinh^{-1}(\sqrt{x}) + C.

Now we're supposed to let y = 2\sinh^{-1}(\sqrt{x}) + C, find the inverse, then "use implicit differentiation to prove that the fundamental theorem of calculus holds, i.e. prove that \frac{dy}{dx} = \frac{1}{\sqrt{(x+x^2)}}". I can find the inverse but I'm not sure how I can use implicit differentiation to prove that. Anything just to get me started would be much appreciated.

 

[Infrared]

I think they want you to find \frac{d}{dx}2sinh^{-1}\sqrt{x} by saying that if y=2sinh^{-1}\sqrt{x} then sinh\frac{y}{2}=\sqrt{x}, which you can differentiate implicitly.

 

[vesu]

I think they want you to find \frac{d}{dx}2sinh^{-1}\sqrt{x} by saying that if y=2sinh^{-1}\sqrt{x} then sinh\frac{y}{2}=\sqrt{x}, which you can differentiate implicitly.

Oh right, that makes sense. Thanks!