On the x-axis and at a distance x from the origin, the gravitational field due to a mass distribution is given by  Axx2+a232 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 :

Given Ex=Axx2+a23/2∫v=∞v=vxdv=-∫∞xEdx∵E=dvdrV(x)-V(∞)=-∫∞xAxx2+a23/2dx   ----(1)Let x2+a2=zdifferentiate bothsides with respective x we get2xdx+0=dz⇒xdx=dz2----(2)From (1) and (2) Given V(∞)=0⇒V(x)-0=-A2∫∞xdzz3/2=-A2∫∞xz-32dz=-A2z-3/2+1-3/2+1=-A2x2+a2-1/2-12∞x=Ax2+a2-1/2