Find F. f '(t) = (4) / 1 + t^2 f(1) = 0

Question

f '(t) = (4) / 1 + t^2 f(1) = 0

Please show all work and answer

Answer 1

f(t) = 4 ∫[1 / (1+t²)]dt = 4Tan^-1t + C

Since f(1) = 0, 4Tan^-1(1) + C = 0

4(π/4) + C = 0

C = -π

So, f(t) = 4Tan^-1t - π