homework and exercises - How do I get the negative sign in the expression for Gravitational Potential Energy?

From universal law of gravitation, gravitational force exerted on a body of mass m by another body of mass M is $$\mathbf F = \frac{GMm}{x^2}$$ where x is the distance between the centres of both the objects.

So, work done by gravitational force in bringing the object of mass m from infinity to a distance r from the centre of body of mass M is $$W = \int_\infty^r \vec{F(x)}.\vec{dx}$$ $$=\int_\infty^r \frac{GMm}{x^2}\hat x.\vec{dx}$$ (where $\hat x$ is the unit vector in the direction in which the body of mass M is attracting the body of mass m, i.e. the direction of $\vec{dx}$ which results the angle between both vectors $0$) $$=\int_\infty^r \frac{GMm}{x^2} {dx}\ cos0$$

$$= - GMm\left(\frac{1}{r}-\frac{1}{\infty}\right)$$ $$= -\frac{GMm}{r}$$

Now, we know that $$W=-(∆U)$$ $$-\frac{GMm}{r} = -(U_r - U_\infty)$$ $$-\frac{GMm}{r} = (U_\infty - U_r)$$ Since, Zero of potential energy is at infinity by convention, so $U_\infty$ = 0 $$-\frac{GMm}{r} = -U_r$$ $$\frac{GMm}{r} = U_r$$

I get potential energy at a distance r as positive, then why is it that gravitational potential energy is $$-\frac{GMm}{r}$$

What is wrong in my derivation?