determine f if f'(x) = (x^2 + 2x)/(sqrt(x^3 + 3x^2+1) and f(0) = 1

Answer 1

Let u = x³ + 3x² + 1 then du/dx = 3x² + 6x or dx = du/(3x² + 6x) = du/(3(x² + 2x))

So the integral becomes ∫(x² + 2x)/(√u•3(x² + 2x)) du = 1/3∫u^-1/2 du = 2/3u^1/2 + C

Back substituting, we get f(x) = 2/3(x³ + 3x² + 1)^1/2 + C

Using f(0) = 1, we can solve for C:

1 = 2/3((0³ + 3(0)² + 1)^1/2 + C

1 = 2/3(1) + C

C = 1/3

f(x) = 2/3(x³ + 3x² + 1)^1/2 + 1/3