The value of the integral $\int_0^2\left(\sqrt{1+x^3}+\sqrt[3]{x^2+2x}\:\right)dx$

Hint. The function $ x \mapsto f(x):=\sqrt{1+x^3}$ is strictly increasing on $[0,2]$, then you may use the following property:

$$ \int_a^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=b{f(b)}-a{f(a)} \tag1 $$

(here $ x \mapsto f^{-1}(x+1)=\sqrt[3]{(x+1)^2-1}=\sqrt[3]{x^2+2x},\quad f^{-1}(0+1)=0,\,f^{-1}(2+1)=2$) obtaining

$$I=2\sqrt{1+2^3}-0\sqrt{1+0^3}=\color{red}{6}$$

as suggested.

Let $f(x)=\sqrt{1+x^3}$. $\\$ Easily show that $f^{-1}(x+1)=\sqrt[3]{x^2+2x}$.

You are asked to find $$\int_0^2 f(x)dx +\int_0^2 f^{-1}(x+1)dx \\ =\int_0^2f(x)dx+\int_1^3 f^{-1}(x)dx \\ =\int_0^2f(x)dx+\int_{f(0)}^{f(2)}f^{-1}(x)dx$$. Draw a picture.

I have one more way to do this:

Let $$I_1=\int_{0}^{2} \sqrt{x^3+1} \:dx$$ By Integration by parts we get

$$I_1=x\sqrt{x^3+1} \bigg|_{0}^{2}-\int_{0}^{2}\frac{3x^3 dx}{2\sqrt{x^3+1}} \tag{1}$$

Now let $$I_2=\int_{0}^{2} (x^2+2x)^{\frac{1}{3}}dx$$ Use substitution $x^2+2x=t^3$ we get

$(2x+2)dx=3t^2dt$ that is

$dx=\frac{3t^2dt}{2(x+1)}$

But $(x+1)^2-1=t^3$ $\implies$

$x+1=\sqrt{t^3+1}$

Also the limits will be again $0$ and $2$

Thus

$$I_2=\int_{0}^{2}\frac{3t^3dt}{2(x+1)}=\int_{0}^{2}\frac{3t^3dt}{2\sqrt{t^3+1}}$$ since $t$ is Dummy variable

$$I_2=\int_{0}^{2}\frac{3x^3dx}{2\sqrt{x^3+1}} \tag{2}$$

Adding $(1)$ and $(2)$ we get

$$I_1+I_2=x\sqrt{x^3+1} \bigg|_{0}^{2}=6$$