Integrating $\int_0^{2} \left(1+x^3\right)^{1/2}+\left(x^2+2x\right)^{1/3} dx$

Question

I stumbled upon this question while solving an exercise of DI.

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

I observed something...

If we denote $f(x)=(1+x^3)^{1/2}$, then the second part of the integrand is merely $f^{-1}(x+1)$ where $f^{-1}(x)$ denotes the inverse of function $f(x)$.

(Had it been something like $\int_{a_1}^{a_2}f(x)dx +\int_{f(a_1)}^{f(a_2)} f^{-1}(x)dx$, It would've been pretty straightforward)

I'm not really sure how should I proceed further...

Any hints would be appreciated

Thanks!

Answer 1

You are almost there. Let $f(x) =\sqrt{1+x^3}$. Notice that $f(0)=1$ and that $f(2)=3$. Also notice that $f^{-1}(x)=\sqrt[3]{x^2-1}$. Your integral then becomes \begin{gather*} \int_0^2 f(x)dx +\int_0^2f^{-1}(x+1)dx = \int_0^2 f(x)dx + \int_1^3 f^{-1}(x)dx = \\ = \int_0^2 f(x)dx + \int_{f(0)}^{f(2)} f^{-1}(x)dx \end{gather*}