Замена √(3x+1) = t, 3x+1 = t^2, x = (t^2-1)/3, dx = 2t/3 dt, t(0) = 1, t(5) = 4
Int(1, 4) 2t/3*1/(2*(t^2-1)/3 +t) dt = 2/3*3*Int(1, 4) t/(2t^2-2+3t) dt =
По методу неопределенных коэффициентов разложим на сумму дробей
t/[(t+2)(2t-1)] = A1/(t+2) + A2/(2t-1) = [A1*(2t-1) + A2*(t+2)]
/[(t+2)(2t-1)]
=
= [t*(2A1 + A2) + (-A1 + 2A2)] /[(t+2)(2t-1)]
Система
{ 2A1 + A2 = 1
{ -A1 + 2A2 = 0
{ 2A1 + A2 = 1
{ -2A1 + 4A2 = 0
Складываем уравнения
5A2 = 1, A2 = 1/5, A1 = 2A2 = 2/5
Интеграл
2/3*3*Int(1, 4) t/(2t^2-2+3t) dt = 2*Int(1, 4) [2/5*1/(t+2) + 1/5*1/(2t+1)] dt =
= 4/5*ln|t+2| + 2/5*1/2*ln|2t+1| |(1, 4) = 4/5*(ln 6 - ln 3) + 2/5*(ln 9 - ln 3) =
= 4/5*ln 2 - 2/5*ln 3