Question

`\rm \int_(0)^2 \left( \sqrt[3]{ {x}^(2) + 2x} + \sqrt{1 + {x}^3 } \right )dx \\`​

Answer 1

If f(x) has an inverse on [a, b], then integrating by parts (take u = f(x) and dv = dx), we can show

`\displaystyle \int_a^b f(x) \, dx = b f(b) - a f(a) - \int_(f(a))^(f(b)) f^(-1)(x) \, dx`

Let
`f(x) = √(1 + x^3)`. Compute the inverse:

`f\left(f^(-1)(x)\right) = \sqrt{1 + f^(-1)(x)^3} = x \implies f^(-1)(x) = \sqrt[3]{x^2-1}`

and we immediately notice that
`f^(-1)(x+1)=\sqrt[3]{x^2+2x}`.

So, we can write the given integral as

`\displaystyle \int_0^2 f^(-1)(x+1) + f(x) \, dx`

Splitting up terms and replacing
`x \to x-1` in the first integral, we get

`\displaystyle \int_1^3 f^(-1)(x) \, dx + \int_0^2 f(x) \, dx = 2 f(2) - 0 f(0) = 2*3+0*1=\boxed{6}`