If f(x) has an inverse on [a, b], then integrating by parts (take u = f(x) and dv = dx), we can show
[tex]\displaystyle \int_a^b f(x) \, dx = b f(b) - a f(a) - \int_{f(a)}^{f(b)} f^{-1}(x) \, dx[/tex]
Let [tex]f(x) = \sqrt{1 + x^3}[/tex]. Compute the inverse:
[tex]f\left(f^{-1}(x)\right) = \sqrt{1 + f^{-1}(x)^3} = x \implies f^{-1}(x) = \sqrt[3]{x^2-1}[/tex]
and we immediately notice that [tex]f^{-1}(x+1)=\sqrt[3]{x^2+2x}[/tex].
So, we can write the given integral as
[tex]\displaystyle \int_0^2 f^{-1}(x+1) + f(x) \, dx[/tex]
Splitting up terms and replacing [tex]x \to x-1[/tex] in the first integral, we get
[tex]\displaystyle \int_1^3 f^{-1}(x) \, dx + \int_0^2 f(x) \, dx = 2 f(2) - 0 f(0) = 2\times3+0\times1=\boxed{6}[/tex]
