Compute $\int _{\frac{4}{5}}^2\:f^{-1}\left(x\right)dx$ where $f\left(x\right)=\frac{-x^3+2x^2-5x+8}{x^2+4},\:x\in \mathbb{R}$ is bijection. We have to compute $\int _{\frac{4}{5}}^2\:f^{-1}\left(x\right)dx$  where $f\left(x\right)=\frac{-x^3+2x^2-5x+8}{x^2+4},\:x\in \mathbb{R}$ is an bijective function.
What we can extract from bijective information?

EDIT : Based on Answers, I tried to Plot this

 A: Hint: If the function bijective, you can draw the function as I drew below
Step 1: Find $a$  and $b$ 
$f(a)=\frac{4}{5} $
$f(b)=2 $
And compute $B=\int _{a}^b\:f\left(x\right)dx$

Step 2:
Compute rectangle area of C and compute $B-C$ 

Step 3:
$A=\int _{\frac{4}{5}}^2\:f^{-1}\left(x\right)dx$
$A+B-C$ is also a rectangle  .
A: You may notice that $f(x)$ is a decreasing function on the interval $[0,1]$ and $f(0)=2,\,f(1)=\frac{4}{5}$, from which it follows that (just look at the picture below to figure it out):
$$ \int_{\frac{4}{5}}^{2}f^{-1}(x)\,dx = -\frac{4}{5}+\int_{0}^{1}f(x)\,dx=\color{red}{\frac{7}{10}+\log 2-\frac{\log 5}{2}}\tag{1}$$
where the last result follows from polynomial division.

A: Let $f(u)=x$, then integration by parts gives
$$
\begin{align}
\int_{4/5}^2f^{-1}(x)\,\mathrm{d}x
&=\int_1^0u\,\mathrm{d}f(u)\\
&=-\int_0^1u\,\mathrm{d}f(u)\\
&=-\left[uf(u)\vphantom{\int}\right]_0^1+\int_0^1f(u)\,\mathrm{d}u\\
&=-\frac45+\int_0^1\frac{-u^3+2u^2-5u+8}{u^2+4}\,\mathrm{d}u\\
&=-\frac45+\int_0^1\frac{2u^2+8}{u^2+4}\,\mathrm{d}u-\int_0^1\frac{u^2+5}{u^2+4}u\,\mathrm{d}u\\
&=\frac65-\frac12\int_4^5\frac{v+1}v\,\mathrm{d}v\\
&=\frac7{10}-\frac12\log\left(\frac54\right)
\end{align}
$$
where $v=u^2+4$.
A: draw a graph of $y = f(x).$  you will see that it has $y$-intercept at $(0,1)$ and has a point $(1,4/5).$  the area you are after is the triangular region bounded by $x = 0$ on the left, by $y = \frac 45$ at the bottom and $y = f(x)$ like a hypotenuse. the area $$\int_{4/5}^2 x\,dy$$ is like cutting the triangle in slices/strips parallel to the $x-$ axis. now, we can find the same area by slicing parallel to the $y$-axis. this way we get the area to be $$ \int_0^1\left(y - \frac 45\right)\, dx$$ 
this is what Mathlover was trying to show you with a picture. hope this is of some use to you.
