Partial Fractions(Integral)

Integral t + 3 / 2t^2 + 3t + 1 dt
My work:
t + 3 / ((2t+1)(t+1)) = A / (2t+1) + B / (t+1)
t + 3 = A(t+1) + B(2t+1)
t + 3 = At + A +2Bt +B
part(a)
1t = At + 2Bt
1 = A + 2B
A = (1-2B)
part(b)
3 = A + B
Sub A = (1 -2B) from part(a)
3 = (1-2B) + B
2 = -B
B = -2
Sub b = -2 in part(a)
A = (1 - 2(-2))
A = 5
So,
Integral 5 / (2t+1) + -2 / (t+1) dt
= 5 In |2t+1| - 2 ln |t+1|
*but the correct answer should be 5/2 In |2t+1| - 2 ln |t+1|