A function has the derivative F'(x) = 6x²/(x³ + 2) for x ≥ 0, and F(0) = 0…

Step 1. Substitute. Let $u = x^3 + 2$. Then $du = 3x^2\,dx$, so $6x^2\,dx = 2\,du$. The integral is $2\displaystyle\int \dfrac{du}{u} = 2\ln|u| + C$.

Step 2. The constant. Since $x^3 + 2 > 0$, $F(x) = 2\ln(x^3 + 2) + C$. At $x = 0$: $0 = 2\ln 2 + C$, so $C = -2\ln 2 = -1.3863$.

Step 3. Evaluate. $F(2) = 2\ln 10 - 2\ln 2 = 2\ln 5 = 3.2189$.