Integral of t^2/(t^2+b^2)^3/2 dt
use integration by parts, with
u = t
dv = t/(t^2+b^2)^3/2 dt
and everything drops out nicely, and you end up with
sinh-1(t/b) - t/√(t^2+b^2) + C
u = t
dv = t/(t^2+b^2)^3/2 dt
and everything drops out nicely, and you end up with
sinh-1(t/b) - t/√(t^2+b^2) + C