Evaluate $\int_0^2 (1+x^3)^{1/2} + (x^2+2x)^{1/3}dx$ 
$$\int_0^2 (1+x^3)^{1/2} + (x^2+2x)^{1/3}dx$$

I figured out that the functions are inverses of each other (kinda). If find the inverse of 1st function, we get
$y=(1+x^3)^{1/2}$
$y^2=1+x^3$
$(y^2-1)^{1/3} = x$
But the function given is, $((x+1)^2-1)^{1/3}$
Substituting $x+1$ would also change the limits. So that is not possible. I don't know what to do here now.
Any help would be appreciated.

By Integral of Inverse Functions we have:
$$\int_c^d f^{-1}(y)\,dy+\int_a^b f(x)\,dx=bd-ac$$
given that $f(a)=c, f(b)=d$.
Let $f(x)= (1+x^3)^{1/2}$. As you have observed, $f^{-1}(y) = (y^2-1)^{1/3}$.
Also we have $f(0)=1$ and $f(2)=3$. Thus:
$$\int_1^3 (y^2-1)^{1/3}\,dy + \int_0^2(1+x^3)^{1/2}\,dx = 3\times2-1\times0=6$$
Do you see how $\displaystyle\int_1^3 (y^2-1)^{1/3}\,dy = \int_0^2 ((x+1)^2-1)^{1/3}\, dx$?
