Answer
From the transform definition, we have
$$\mathcal{L}[m t]=\lim _{T \rightarrow \infty}\left[\int_{0}^{T} m t e^{-s t} d t\right]=m \lim _{T \rightarrow \infty}\left[\int_{0}^{T} t e^{-s t} d t\right]$$
The method of integration by parts states that
$$\int_{0}^{T} u d v=\left.u v\right|_{0} ^{T}-\int_{0}^{T} v d u$$
Choosing $u=t$ and $d v=e^{-s t} d t,$ we have $d u=d t, v=-e^{-s t} / s,$ and
$$\mathcal{L}[m t]=m \lim _{T \rightarrow \infty}\left[\int_{0}^{T} t e^{-s t} d t\right]=m \lim _{T \rightarrow \infty}\left[\left.t \frac{e^{-s t}}{-s}\right|_{0} ^{T}-\int_{0}^{T} \frac{e^{-s t}}{-s} d t\right]$$
$$=m \lim _{T \rightarrow \infty}\left[\left. t \frac{e^{-s t}}{-s} \right|_{0} ^{T}-\left.\frac{e^{-s t}}{(-s)^{2}}\right|_ { 0 } ^ { T }\right] $$
$$=m \lim _{T \rightarrow \infty}\left[\frac{T e^{-s T}}{-s}-0-\frac{e^{-s T}}{(-s)^{2}}+\frac{e^{0}}{(-s)^{2}}\right]=\frac{m}{s^{2}}$$
because, if we choose the real part of $s$ to be positive, then
$$\lim _{T \rightarrow \infty} T e^{-s T}=0$$
Work Step by Step
From the transform definition, we have
$$\mathcal{L}[m t]=\lim _{T \rightarrow \infty}\left[\int_{0}^{T} m t e^{-s t} d t\right]=m \lim _{T \rightarrow \infty}\left[\int_{0}^{T} t e^{-s t} d t\right]$$
The method of integration by parts states that
$$\int_{0}^{T} u d v=\left.u v\right|_{0} ^{T}-\int_{0}^{T} v d u$$
Choosing $u=t$ and $d v=e^{-s t} d t,$ we have $d u=d t, v=-e^{-s t} / s,$ and
$$\mathcal{L}[m t]=m \lim _{T \rightarrow \infty}\left[\int_{0}^{T} t e^{-s t} d t\right]=m \lim _{T \rightarrow \infty}\left[\left.t \frac{e^{-s t}}{-s}\right|_{0} ^{T}-\int_{0}^{T} \frac{e^{-s t}}{-s} d t\right]$$
$$=m \lim _{T \rightarrow \infty}\left[\left. t \frac{e^{-s t}}{-s} \right|_{0} ^{T}-\left.\frac{e^{-s t}}{(-s)^{2}}\right|_ { 0 } ^ { T }\right] $$
$$=m \lim _{T \rightarrow \infty}\left[\frac{T e^{-s T}}{-s}-0-\frac{e^{-s T}}{(-s)^{2}}+\frac{e^{0}}{(-s)^{2}}\right]=\frac{m}{s^{2}}$$
because, if we choose the real part of $s$ to be positive, then
$$\lim _{T \rightarrow \infty} T e^{-s T}=0$$
