On the $$x$$-axis and at a distance $$x$$ from the origin, the gravitational field due to a mass distribution is given by $$\frac{Ax}{(x^2 + a^2)^{3/2}}$$ in the $$x$$-direction. The magnitude of the gravitational potential on the $$x$$-axis at a distance $$x$$, taking its value to be zero at infinity is:

We are told that on the $$x$$-axis the gravitational field (which we may denote by $$\vec g(x)$$) has only an $$x$$-component and is given by

$$\vec g(x)=\frac{A x}{(x^2+a^2)^{3/2}}\;\hat i.$$

For a one-dimensional situation along the $$x$$-axis, the relation between gravitational field and gravitational potential $$\phi(x)$$ is the well-known formula

$$\vec g(x)= -\,\frac{d\phi}{dx}\;\hat i.$$

In scalar form (the direction is already along $$\hat i$$) this becomes

$$\frac{d\phi}{dx}= -\,\frac{A x}{(x^2+a^2)^{3/2}}.$$

We now integrate both sides with respect to $$x$$. Taking the potential to be zero at infinity, we write

$$\phi(x)-\phi(\infty)=\int_{\infty}^{x}\frac{d\phi}{dx'}\,dx'=-\int_{\infty}^{x}\frac{A x'}{(x'^2+a^2)^{3/2}}\,dx'.$$

Because $$\phi(\infty)=0$$, this simplifies to

$$\phi(x)= -A\int_{\infty}^{x}\frac{x'}{(x'^2+a^2)^{3/2}}\;dx'.$$

To evaluate the integral, we first consider the indefinite form

$$\int \frac{x}{(x^2+a^2)^{3/2}}\;dx.$$

We observe that

$$\frac{d}{dx}\left(\frac{1}{\sqrt{x^2+a^2}}\right)=\frac{d}{dx}\left((x^2+a^2)^{-1/2}\right)= -\frac{1}{2}(x^2+a^2)^{-3/2}\cdot 2x=-\frac{x}{(x^2+a^2)^{3/2}}.$$

Thus,

$$\int \frac{x}{(x^2+a^2)^{3/2}}\;dx = -\frac{1}{\sqrt{x^2+a^2}}+C.$$

Substituting this result back into the definite integral we have

$$\phi(x)= -A\Bigl[-\frac{1}{\sqrt{x'^2+a^2}}\Bigr]_{\infty}^{x}=A\Bigl[\frac{1}{\sqrt{x'^2+a^2}}\Bigr]_{\infty}^{x}.$$

Now we evaluate the limits. At the upper limit $$x' = x$$ we obtain $$\dfrac{1}{\sqrt{x^2+a^2}}$$, while at the lower limit $$x' \to \infty$$ we have $$\dfrac{1}{\sqrt{\infty}}=0$$. Hence

$$\phi(x)=\frac{A}{\sqrt{x^2+a^2}}.$$

Therefore the magnitude of the gravitational potential on the $$x$$-axis at a distance $$x$$ is

$$\boxed{\displaystyle \frac{A}{\sqrt{x^2+a^2}}}.$$

Hence, the correct answer is Option A.