solve the initial value problem

Question

f’(x) = 1/(x^3) + 1/(1+x^2)

f(1) = 33

I got up to the antiderivative of -1/(2x^2) + tan^-1 (x) + c

Answer 1

f’(x) = 1/(x^3) + 1/(1+x^2); f(1) = 33

In order to solve this initial value problem (IVP), we must integrate f'(x) and then plug in the initial condition to solve for the constant to determine f(x):

f’(x) = 1/(x^3) + 1/(1+x^2)

∫f’(x)dx = ∫[1/(x^3) + 1/(1+x^2)]dx

f(x) = ∫[x^-3 + 1/(1+x^2)]dx

f(x) = (-1/2)x^-2 + arctan(x) + C

f(1) = (-1/2)(1)^-2 + arctan(1) + C

33 = -1/2 + π/4 + C

67/2 - π/4 = C

f(x) = (-1/2)x^-2 + arctan(x) + 67/2 - π/4

f(x) = -1/(2x²) + arctan(x) + 67/2 - π/4

Hope this helped! Recall the known derivative d/dx[1/(1+x²)] = arctan(x).

Answer 2

f(x) = - 1/(2x²) + tan^-1x + C; f(1) = - 1/2 + π/4 + C = 33; C = 33.5 - π/4; f(x) = - 1/(2x²) + tan^-1x + 33.5 - π/4