Answer
\[\frac{1}{36}\]
Work Step by Step
Let \[I=\int_{1}^{\infty}\frac{1}{(2x+1)^3}dx\]
This is improper integral
\[\Rightarrow I=\lim_{t\rightarrow \infty}\int_{1}^{t}\frac{1}{(2x+1)^3}dx \;\;\;\ldots (1)\]
\[I=\lim_{t\rightarrow \infty}\int_{1}^{t}(2x+1)^{-3}dx\]
\[I=\lim_{t\rightarrow \infty}\left[\frac{(2x+1)^{-2}}{-2(2)}\right]_{1}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[\frac{-1}{4(2t+1)^2}+\frac{1}{4(3)^2}\right]\]
\[I=\frac{1}{36}\]
Since limit on R.H.S. of (1) exists so $I$ is convergent and $I=\frac{1}{36}$.
