Help with calculus 12 hw!

I can't figure this question out, someone please help!
Solve the initial value problem dy/dx = 1+x(x^3/2) y(0)=1

dy/dx = 1 + x(x^3/2) = 1 + x^(5/2)
==> dy = 1 + x^(5/2) dx
==> ∫ dy = ∫ 1 + x^(5/2) dx
==> y = x + (2/7)x^(7/2) + C

Plug in the point y(0) = 1 to solve for C:
1 = 0 + 0 + C ==> C = 1

So:
y = x + (2/7)x^(7/2) + 1

Integral of (1 + x(x^3/2))dx = Integral of (1 + x^(5/2))dx = x + 2/7 x ^(7/2) + C

y(x=0) = 1 = 0 + 2/7 0^(7/2) + C means C = 1.