tentukan persamaan kurva jika diketahui fungsi turunannya dy/dx=2/(2x+1)3 dan melalui titik (-1,1/3)

dy/dx = y' = 2/(2x + 1)³

y = f(x) = ∫y' dx
f(x) = ∫2/(2x + 1)³ dx

Misal :
u = 2x + 1
du/dx = 2
2 dx = du

f(x) = ∫u^-3 . du
f(x) = -1/2 . u^-2 + C
f(x) = -1 / (2(2x + 1)²) + C

Mlalui (-1 , 1/3)
f(-1) = 1/3
-1 / 2(2(-1) + 1)² + C = 1/3
-1/2 + C = 1/3
C = 1/3 + 1/2
C = 5/6

Persamaan Kurva :
f(x) = -1/(2(2x + 1)²) + C
f(x) = -1/(2(2x + 1)²) + 5/6