Answer
$$
\int_{1}^{\infty} \frac{1}{(2 x+1)^{3}} d x=\frac{1}{36}
$$
Work Step by Step
The given integral
$$
\int_{1}^{\infty} \frac{1}{(2 x+1)^{3}} d x
$$
is an improper integral, hence
$$
\begin{aligned} \int_{1}^{\infty} \frac{1}{(2 x+1)^{3}} d x &=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1}{(2 x+1)^{3}} d x \\
&=\lim _{t \rightarrow \infty} \int_{1}^{t} \frac{1}{2}(2 x+1)^{-3} 2 d x\\
&=\lim _{t \rightarrow \infty}\left[-\frac{1}{4(2 x+1)^{2}}\right]_{1}^{t} \\ &=-\frac{1}{4} \lim _{t \rightarrow \infty}\left[\frac{1}{(2 t+1)^{2}}-\frac{1}{9}\right] \\
&=-\frac{1}{4}\left(0-\frac{1}{9}\right) \\
&=\frac{1}{36} \end{aligned}
$$
