$\underset{0}{\overset{\infty }{\int }}\frac{45a}{{(3+a+at)}^4}dt=$

$\underset{0}{\overset{\infty }{\int }}\frac{45a}{{(3+a+at)}^4}dt$

$=\underset{3+a}{\overset{\infty }{\int }}\frac{45dy}{y^4}$

$$\begin{gathered} 3+a+at=y \\ \Rightarrow a dt=dy; \\ t=0\Rightarrow y=3+a \\ t\to \infty \Rightarrow y\to \infty \end{gathered}$$

$$\begin{gathered} =\left(45\right){\left(\frac{y^{-3}}{-3}\right)}_{3+a}^{\infty } \\ =15{\left[-\frac{1}{y^3}\right]}_{3+a}^{\infty } \\ =15{\left[\frac{-1}{y^3}\right]}_{3+a}^{\infty } \\ =15\left[\frac{1}{{\left(3+a\right)}^3}-0\right]=\frac{15}{{\left(3+a\right)}^3} \end{gathered}$$