[tex]\mathcal L_s\{t\cosh3t\}=\displaystyle\int_0^\infty t\cosh3t e^{-st}\,\mathrm dt[/tex]
Integrate by parts, setting
[tex]u_1=t\implies\mathrm du_1=\mathrm dt[/tex]
[tex]\mathrm dv_1=\cosh3t e^{-st}\,\mathrm dt\implies v_1=\displaystyle\int\cosh3t e^{-st}\,\mathrm dt[/tex]
To evaluate [tex]v_1[/tex], integrate by parts again, this time setting
[tex]u_2=\cosh3t\implies\mathrm du_2=3\sinh3t\,\mathrm dt[/tex]
[tex]\mathrm dv_2=\displaystyle\int e^{-st}\,\mathrm dt\implies v_2=-\frac1se^{-st}[/tex]
[tex]\implies\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}+\frac3s\int \sinh3te^{-st}[/tex]
Integrate by parts yet again, with
[tex]u_3=\sinh3t\implies\mathrm du_3=3\cosh3t\,\mathrm dt[/tex]
[tex]\mathrm dv_3=e^{-st}\,\mathrm dt\implies v_3=-\dfrac1se^{-st}[/tex]
[tex]\implies\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}+\frac3s\left(-\frac1s\sinh3te^{-st}+\frac3s\int\cosh3te^{-st}\,\mathrm dt\right)[/tex]
[tex]\displaystyle\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}-\frac3{s^2}\sinh3te^{-st}+\frac9{s^2}\int\cosh3te^{-st}\,\mathrm dt[/tex]
[tex]\displaystyle\frac{s^2-9}{s^2}\int\cosh3te^{-st}\,\mathrm dt=-\frac1s\cosh3te^{-st}-\frac3{s^2}\sinh3te^{-st}[/tex]
[tex]\implies\displaystyle\underbrace{\int\cosh3te^{-st}\,\mathrm dt}_{v_1}=-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}[/tex]
So we have
[tex]\displaystyle\int_0^\infty t\cosh3t e^{-st}\,\mathrm dt=u_1v_1\big|_{t=0}^{t\to\infty}-\int_0^\infty v_1\,\mathrm du_1[/tex]
[tex]=\displaystyle-\frac{t(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\bigg|_{t=0}^{t\to\infty}-\int_0^\infty \left(-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\right)\,\mathrm dt[/tex]
[tex]=\displaystyle\frac1{s^2-9}\int_0^\infty(s\cosh3t+3\sinh3t)e^{-st}\,\mathrm dt[/tex]
We already have the antiderivative for the first term:
[tex]\displaystyle\frac s{s^2-9}\int_0^\infty \cosh3te^{-st}\,\mathrm dt=\frac s{s^2-9}\left(-\frac{(s\cosh3t+3\sinh3t)e^{-st}}{s^2-9}\right)\bigg|_{t=0}^{t\to\infty}[/tex]
[tex]=\dfrac{s^2}{(s^2-9)^2}[/tex]
And we can easily find the remaining term's antiderivative by integrating by parts (for the last time!), or by simply exchanging [tex]\cosh[/tex] with [tex]\sinh[/tex] in the derivation of [tex]v_1[/tex], so that we have
[tex]\displaystyle\frac3{s^2-9}\int_0^\infty\sinh3te^{-st}\,\mathrm dt=\frac3{s^2-9}\left(-\frac{(s\sinh3t+3\cosh3t)e^{-st}}{s^2-9}\right)\bigg|_{t=0}^{t\to\infty}[/tex]
[tex]=\dfrac9{(s^2-9)^2}[/tex]
(The exchanging is permissible because [tex](\sinh x)'=\cosh x[/tex] and [tex](\cosh x)'=\sinh x[/tex]; there are no alternating signs to account for.)
And so we conclude that
[tex]\mathcal L_s\{t\cosh3t\}=\dfrac{s^2+9}{(s^2-9)^2}[/tex]
