dv/dt = 7/(1+t^2) +(sect)^2
dv = 7dt/(1+t^2) +(sect)^2 dt
v = 7∫dt/(1+t^2) +∫(sect)^2 dt
The first integral can be solved with the arctan trig substitution. The second integral can be solved recalling that tan(t) is the antiderivative of sec2(t).
v(t) = 7arctan(t) + tan(t) + v0
Apply the initial condition, v(0) = 3. Let t = 0, then
7arctan(0) + tan(0) + v0 = 3, which implies that v0 = 3.
Thus, we have
v(t) = 7arctan(t) + tan(t) + 3