How do you obtain this electric potential for a nonconducting rod: V = ak [ (L+ d) ln ( 1 + l/d) - L)

It has a negligible thickness, located along the x-axis. Its ends have coordinates x= L and x=0. It also has a linear charge density λ= ax; a is just a constant. At an infinite distance, V =0. How do I show that the electric potential is V = ak [ (L+ d) ln ( 1 + l/d) - L)

for the location x= L + d.

Attempt:
V= integral of dq/r. I think r=x.


dq= λ dx.

dq= ax dr.

I have given this problem a few hours, and I need some feedback.

Is a sum/difference of integrals involved?

* 3 hours ago
* - 4 days left to answer.

Additional Details
I'm able to obtain " ak [ (L+ d) ln ( 1 + l/d)" so far. But I don't know how " - L" is involved." "I'm missing that part.

2 minutes ago

You've got the right idea, but r is not just x. Draw a sketch with the rod on the x-axis with one end at the origin the other end at "L".
Choose a "dq" located at "x" . Its potential at L+d is;
dV = kdq/r = kdq/(L+d-x) = kaxdx/(L+d-x)

Integrate that from x=0 to x=L .

BTW your solution looks a little funny. I believe the argument of Ln should be (1+L/d) ; Not (1+1/d).