f'(t) = 12 / (1+t^2) , f(1)=0 . Find f.

Answer 1

f'(x)= 12/(1+t^2)= 12(1+t^2)^-1

f(x)= 12arctanx + C

f(1) =0 = 12arctan1 +C

C= -12arctan1

f(x)= 12arctanx -12arctan1

Answer 2

Need to integrate to find f(t)

The integral of 1/(1+t²) is a standard integral, tan^-1(t) , so just need to multiply by 12 and add the constant, C, giving

f(t)=12tan^-1(t) + C

to solve for C:

f(1) = 12tan^-1(1) + C = 0

C = -12tan^-1(1)

If you need to prove 1/(1+t²) = tan^-1(t), let t = tan(u)