Find the potential a distance z above the center of a uniform line of charge with charge density lambda and length 2L .

For a continuous potential to equation is
V=int (k*dq)/r
Where k=1/(4pi epsilon_0) , dq is an infinitesimal piece of charge, and r is the distance away dq is from the point we are measuring.
We know the charge per unit length lambda=(dq)/(dl) where dl is an infinitesimal chuck of the wire. If we let the wire lay along the x-axis then dl=dx . Therefore,
dq=lambda dx
Now, looking at the picture we have a right triangle. So we know r=sqrt(x^2+z^2) . Then the integral becomes,
V=int_(-L)^L (k*lambda dx)/sqrt(x^2+z^2)=k*lambda int_(-L)^L dx/sqrt(x^2+z^2)
From a table of integrals we can find that
V=k lambda ln(x+sqrt(x^2+z^2))|_(-L)^L
V=lambda/(4pi epsilon_0) ln((L+sqrt(L^2+z^2))/(-L+sqrt(L^2+z^2)))
To simplify multiply the numerator and denominator by L+sqrt(L^2+z^2)
V=lambda/(4pi epsilon_0)ln[(L+sqrt(L^2+z^2))^2/(-L^2+L^2+z^2)]
V=2*lambda/(4pi epsilon_0)ln[(L+sqrt(L^2+z^2))/z]
V=lambda/(2pi epsilon_0)ln[(L+sqrt(L^2+z^2))/z]
This is the answer.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/potlin.html