Mistake in integrating the inverse of a function? The problem I have is:
$$\int_0^{\frac1{\sqrt{3}}}\sqrt{x+\sqrt{x^2+1}}dx$$
I'm not asking how to solve it, I'm asking if anyone can point out to me the error that I have made:
By differentiating both sides, one can easily verify (see edit):
$$\int f^{-1}(x)dx = xf^{-1}(x)-F(f^{-1}(x))$$
Where $F'(x)=f(x)$
I recognized the integrand as being the inverse of $f(x)=\frac12(x^2+x^{-2})$ and therefore we have $$F(x)=\frac12\left(\frac{x^3}3-\frac1x\right)=\frac{x^4-3}{6x}$$
Thus we can simplify the integral to:
$$[xf^{-1}(x)-F(f^{-1}(x))]^{1/\sqrt{3}}_0=\frac{f^{-1}\left(\frac1{\sqrt3}\right)}{\sqrt3}-F\left(f^{-1}\left(\frac1{\sqrt3}\right)\right)+F(f^{-1}(0))$$
Remembering that the integrand is $f^{-1}(x)$, we can plug in values to get:
$$f^{-1}(0)=1$$
$$f^{-1}\left(\frac1{\sqrt3}\right)=3^{1/4}$$
And thus the integral can be reduced to:
$$\frac{3^{1/4}}{\sqrt3}-F(3^{1/4})+F(1)$$
By evaluating, we can see that $F(3^{1/4})=0$ and $F(1)=\frac13$ and so it all simplifies to:
$$\frac{3^{1/4}}{\sqrt3}-\frac13=3^{-1/4}-\frac13$$
However Wolfram Alpha tells me that the answer is 2/3. Can someone please tell me where I messed up?
Edit:
For those who doubted the formula that I gave:
If you take the right hand side of the equation and take the derivative, we get:
$$\frac{d}{dx}[xf^{-1}(x)-F(f^{-1}(x))] = \frac{d}{dx}xf^{-1}(x)-\frac{d}{dx}F(f^{-1}(x))$$
$$=x\frac{d}{dx}f^{-1}(x)+f^{-1}(x) -\frac{d}{dx}F(f^{-1}(x))$$
$$=x\frac{d}{dx}f^{-1}(x)+f^{-1}(x) -f(f^{-1}(x))\frac{d}{dx}f^{-1}(x)$$
$$=x\frac{d}{dx}f^{-1}(x)+f^{-1}(x) -x\frac{d}{dx}f^{-1}(x)$$
$$=f^{-1}(x)$$
So therefore:
$$\int f^{-1}(x)dx = xf^{-1}(x)-F(f^{-1}(x))$$

Let $x=f(y)$ and we assume that we can write $y=f^{-1}(x)$.  Then
$$\int dx \, f^{-1}(x) = y f(y) - \int dy \, f(y) = x f^{-1}(x) - F[f^{-1}(x)]$$
as you stated.
$$x=\frac12 \left ( y^2 + y^{-2}\right ) \implies y^4 - 2 x y^2 + 1 =0 \implies y^2=x\pm \sqrt{x^2-1}$$
Your inverse function is wrong.  You actually want
$$x=\frac12 \left ( y^2 - y^{-2}\right ) \implies y^4 - 2 x y^2 - 1 =0 \implies y^2=x\pm \sqrt{x^2+1}$$
Then the indefinite integral is
$$\frac{y}{2} \left ( y^2 - y^{-2}\right )  - \frac12 \int dy \, \left ( y^2 - y^{-2}\right )  = \frac13 y^3 -\frac1{y} + C$$
It is true that when $x=3^{-1/2}$, $y=3^{1/4}$, and when $x=0$, $y=1$.  Then, the definite integral sought is
$$\left [ \frac13 y^3 -\frac1{y} \right ]_{y=3^{1/4}} - \left [ \frac13 y^3 -\frac1{y} \right ]_{y=1}= \frac{2}{3}$$
