Solution) Let length of the cylinder = L

 

Radius of the cylinder be = R

 

Now consider the circular base of cylinder through which the flux due to enclosed charge Q is passing.

 

Our approach will be

 

 

Here Factor 2 is introduced  to take care of bottom as well as top circular bases for a cylinder.

 

Now consider bottom circular base.

Let us consider a ring of inner radius = r

Also, outer raius of the ring be = r + dr  (Where, dr is infinitesimal)

 

Now area of this ring = 2r.dr

 

Also electric field due to charge Q at the center, at every point of the ring = E

 

Here, E = k Q / (L²/4 + r²)

 

So electric flux through the ring =d =  E.da = [ k Q / (L²/4 + r²) ] . 2r. dr

 

Now flux through the entire bottom =  =  d = _{[0 ]}^{[R ]} [k Q / (L²/4 + r²)].2r .dr

 

 

now evaluate this integral and use