1. dy/dx = x^3/2,kurva melalui titik (1,9) 2. dy/dx x^2 - 1/x^2, kurva melalui titik (1,1/3) 3. dy/dx = 1/√x,kurva melalui titik (9,4)

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y= ∫ dy/dx = ∫ x^3/2 dx = 2/5 x^5/2 + c
(x,y) =(1,9)
2/5 (1)^5/2 + c = 9
2/5 (1) + c = 9
c = 9 - 2/5 = 43/5
y= 2/5 x^5/2 + 43/5
y = 1/3 (2x²√ x  + 43)

y= ∫dy.dx = ∫ (x^2 - 1/x^2) dx = ∫(x^2 - x^-2) dx
y = 1/3 x^3 + x^-1 + c
y = 1/3 x^3  + 1/x + c
(x,y)=(1.  1/3)
1/3 = 1/3(1)^3 + 1/1 + c
1/3 = 1/3 + 1 + c
c = - 1
y = 1/3 x^3 + 1/x  - 1

y = ∫ 1/√x dx = ∫ 1/x^1/2 dx = ∫ x^-1/2 dx
y = 2 x^1/2 + c
y = 2√x  + c
(x,y)=(9,4)
4 = 2√9 + c
4 = 2(3) + c
4 = 6 +c
c = -2
y = 2√x - 2