differetiation equation dy/dx = e^y*square root (x+1)

I am not quite sure I have a grasp on this.

dy/dx = e^y*square root (x+1)

dy/dx = e^y(x+1)^1/2
d/dy e^y = d/dx(x+1)^1/2

integral d/dy e^y = integral d/dx (x+1)
ln[y] = 3/2(x+1)^3/2

\(\displaystyle \L\\\frac{dy}{dx}=e^{y}\sqrt{x+1}\)

Separate variables:

\(\displaystyle \L\\\frac{dy}{e^{y}}=\sqrt{x+1}dx\)

Integrate:

\(\displaystyle \L\\\int{e^{-y}}dy=\int\sqrt{x+1}dx\)

\(\displaystyle \L\\-e^{-y}=\frac{2(x+1)^{\frac{3}{2}}}{3}+C\)

See?.