Answer
Convergent$\;,\;\large\frac{1}{18}$
Work Step by Step
Let
\[I=\int_{1}^{\infty}\frac{1}{(2x+1)^3}dx\;\;\;\ldots(1)\]
\[I=\lim_{t\rightarrow \infty}\int_{1}^{t}\frac{1}{(2x+1)^3}dx\;\;\;\ldots(2)\]
\[I=\lim_{t\rightarrow \infty}\left[-\frac{1}{2}(2x+1)^{-2}\right]_{1}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[-\frac{1}{2(2t+1)^2}+\frac{1}{2(3)^2}\right]\]
\[I=\frac{1}{18}\]
Since limit on R.H.S. of (2) exists
So given improper integral (1) is convergent
\[I=\frac{1}{18}.\]
