How do I find the sum of the series

The sum of:

Sum from 0 to infinity

[(3x+1)^(2n)] / [4^n(n+1)]

Can somebody please explain how to find the sum?

Thank you.

Note that we can rewrite this as
Σ(n = 0 to ∞) [(3x+1)^2 / 4]^n / (n+1).

To compute this sum, start with the geometric series
1/(1 - t) = Σ(n = 0 to ∞) t^n.

Integrate both sides from 0 to t:
-ln(1 - t) = Σ(n = 0 to ∞) t^(n+1)/(n+1).

Divide both sides by t:
Σ(n = 0 to ∞) t^n / (n+1) = -ln(1 - t)/t.

Finally, let t = (3x+1)^2 / 4:
Σ(n = 0 to ∞) [(3x+1)^2 / 4]^n / (n+1)
= -ln(1 - (3x+1)^2/4) / ((3x+1)^2 / 4)
= -4 ln(1 - (3x+1)^2/4) / (3x+1)^2.

I hope this helps!