we know
#f'(t)=dy/dt#
#dy/dt=(12)/(1+t^2)#
transposing differentials
#dy=(12)/(1+t^2)dt#
integrating
#int(12)/(1+t^2)dt#
#t=tan(u)#
#dt=1/cos^2(u)du#
#int(12)/((cos^2(u)/(cos^2(u))+(sin^2(u))/(cos^2(u))))(1/cos^2(u)du)#
#cos^2(u)+sin^2(u)=1#
#int(12)/((1/(cos^2(u))))(1/cos^2(u)du)#
dividing
#int(12)du#
integrating
#int(12)du=12u+c#
but #arctg(t)=u#
then
#12u+c=12arctg(t)+c#
#f(t)=12arctg(t)+c# ... 1
now finding the constant 'c' #f(1)=0#
in (1)
#f(1)=12arctg(1)+c#
#0=12arctg(1)+c#
#arctg(1)=Pi/4#
#0=12Pi/4+c#
#0=3Pi+c#
#-3Pi=c#
now
#f(t)=12arctg(t)-3Pi#
