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!