Answer:
It converges to 1/8.
Step-by-step explanation:
We are given the integral: [tex]\int\limits^\infty_0 \frac{dx}{(4x+1)^3}[/tex]
[tex]\int\limits^\infty_0 \frac{dx}{(4x+1)^3}= \lim_{t \to \infty} \int\limits^t_0 \frac{dx}{(4x+1)^3}=\lim_{t \to \infty} \frac{-1}{8(4x+1)^2}|^t_0 = 0-(-\frac{1}{8})=\frac{1}{8}[/tex]
So it is convergent and converges to 1/8.
