An Inverse Integration

Let us take $f(x)=(x^2+2x)^{\frac{1}{3}}$ . Then, we have $f^{-1}(x)=(x^3+1)^{\frac{1}{2}}-1$

So, the integral that we want to find out can be written as follows:

$I=\int \limits_0^2 \left(f^{-1}(x)+1+f(x)\right) \,dx$

Note that $f(0)=0$ and $f(2)=2$ , i.e., $f(a)=a$ and $f(b)=b$ where $a,b$ are the lower and upper limits respectively of the given integral. Also, $f(x)$ and $f^{-1}(x)$ are absolutely continuous on $[0,2]$ .

We will then use the following theorem,

Image

This results in the following:

$I=(2\times 2)-(0\times 0)+\int \limits_0^2 dx = 4+[x]_0^2=4+(2-0)=\boxed{6}$

For more details on the theorem used, refer to this link .

If there's anything wrong with the solution, feel free to notify me. :)

P.S - Special thanks to Ronak Agarwal for pointing me in the right direction.