If ∫(x+1)/√(2x-1) dx = f(x)√(2x-1) + C, where C is a constant of integration, then f(x) is equal to:

The correct option is A
13(x+4)√(2x-1) = t ⇒ 2x-1 = t² ⇒ 2dx = 2t.dt
∫(x+1)/√(2x-1) dx = ∫(t²+1)/2t * t dt = ∫(t²/2 + 1/2) dt = t³/6 + t/2 + C = (2x-1)^(3/2)/6 + √(2x-1)/2 + C = √(2x-1) * [(2x-1)/6 + 1/2] + C = √(2x-1) * [(2x-1+3)/6] + C = √(2x-1) * (2x+2)/6 + C = √(2x-1) * (x+1)/3 + C
Comparing this with f(x)√(2x-1) + C, we get f(x) = (x+1)/3. This is not one of the options. Let's try another approach.
Let I = ∫(x+1)/√(2x-1) dx. Let 2x-1 = t², then 2dx = 2t dt, so dx = t dt. Also, x = (t²+1)/2. Then
I = ∫[(t²+1)/2 + 1]/t * t dt = ∫(t²/2 + 3/2)dt = t³/6 + (3/2)t + C = (2x-1)^(3/2)/6 + (3/2)√(2x-1) + C
= √(2x-1)[(2x-1)/6 + 3/2] + C = √(2x-1)[(2x-1+9)/6] + C = √(2x-1)(2x+8)/6 + C = √(2x-1)(x+4)/3 + C
Comparing this with f(x)√(2x-1) + C, we have f(x) = (x+4)/3. Therefore, the closest option is 1/3(x+4).