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}{{\left(x^2+a^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity, is :

$\begin{array}{l}\text{ Given }{E}_x=\frac{Ax}{{\left(x^2+a^2\right)}^{3/2}}\\ {\int }_{v=\infty }^{v=v_x}dv=-{\int }_{\infty }^{x}Edx\left(\because E=\frac{dv}{dr}\right)\\ V\left(x\right)-V(\infty )=-{\int }_{\infty }^x\frac{Ax}{{\left(x^2+a^2\right)}^{3/2}}dx ----\left(1\right)\\ Let \left(x^2+a^2\right)=z\\ differentiate bothsides with respective x we get\\ 2xdx+0=dz\Rightarrow xdx=\frac{dz}{2}----\left(2\right)\\ From \left(1\right) and \left(2\right)\\ \text{ Given }V(\infty )=0\Rightarrow V\left(x\right)-0=-\frac{A}{2}{\int }_{\infty }^x\frac{dz}{{\left(z\right)}^{3/2}}=-\frac{A}{2}{\int }_{\infty }^x{\left(z\right)}^{-\frac{3}{2}}dz=-\frac{A}{2}\left(\frac{z^{-3/2+1}}{-3/2+1}\right)\\ =-{\left.\frac{A}{2}\frac{{\left(x^2+a^2\right)}^{-1/2}}{\left(-\frac{1}{2}\right)}\right|}_{\infty }^x=A{\left(x^2+a^2\right)}^{-1/2}\end{array}$