Definite integral of the inverse of a function We have $f:\mathbb{R}\to \mathbb{R}$ with $$f\left(x\right)=\frac{\left(-x^3+2x^2-5x+8\right)}{\left(x^2+4\right)}$$
Knowing that the function is bijective, calculate $$\int _{\frac{4}{5}}^2f^{-1}(x)dx$$
How do I solve this? I can't really compute $f^{-1}(x)$. Book tells me to use substitution but I don't really understand what it means.
 A: Make $u=f^{-1}(x) \implies f(u)=x, dx = f'(u)du$. Note that $f^{-1}(4/5)=1, f^{-1}(2)=0$.
$$
\int_{4/5}^{2}f^{-1}(x)dx = \int_{1}^{0}uf'(u)du = -\int_{0}^{1}uf'(u)du
$$
Can you solve it from here?
A: If $y_1 = f(x_1)$ and $y_2 = f(x_2)$, then $$\int_{y_1}^{y_2}f^{-1}(x)\,{\rm d}x = \int_{y_1}^{y_2}f^{-1}(y)\,{\rm d}y = \int_{x_1}^{x_2}xf'(x)\,{\rm d}x,$$because $x = f^{-1}(y) \implies y = f(x) \implies {\rm d}y = f'(x)\,{\rm d}x$. The limits of integration go like this because is $f$ bijective. Now just plug in everything and solve the resulting integral.
A: 
Lemma: If $f(x)$ is a continuous and increasing function and $a<b$, then:
  $$ \int_{a}^{b} f(x)\,dx + \int_{f(a)}^{f(b)}f^{-1}(x)\,dx = b\, f(b)-a\, f(a). $$

To prove it, you just need to draw a picture or look at this Wikipedia page.
In our case, $f(x)$ is decreasing and $f^{-1}(2)=0,\,f^{-1}(4/5)=1$, so the problem boils down to computing:
$$ \int_{0}^{1}f(x)\,dx = \int_{0}^{1}\left(2-x-\frac{x}{4+x^2}\right)\,dx = 2-\frac{1}{2}-\frac{1}{2}\left.\left(\log(4+x^2)\right)\right|_{x=0}^{1}.$$
Can you fill the gaps?
