For the first :
let u = ⁴√(2x+3).
u⁴ = 2x + 3, x = (u⁴ - 3)/2
4u³du = 2dx , so dx =2u³du
I = ∫ [(u⁴ - 3)/2].2u³du / u = ∫ u²(u⁴ - 3).du
For the second :
y = 0, or e^(-1/x).y + x².y' = 0
y' = -(1/x²).e^(-1/x).y
y = µ.e^(-e^(-1/x))
For the third :
(1-x²)y' = x(y+y²)
so :
y' / (y+y²) = x / (1-x²)
...