$ Consider\\[3ex] \displaystyle \int_0^{\infty{}}e^{-st}\frac{\sin {2t+\sin {3t}}}{t}dt=L\left[\frac{\sin {2t+\sin {3t}}}{t}\right] \\[2ex] \displaystyle L\left[\sin {2t}\right]=\ \frac{2}{{s^2+2}^2} \\[2ex] \displaystyle L\left[\sin {3t}\right]=\ \frac{3}{{s^2+3}^2} \\[2ex] \displaystyle If\ \ L\left[f(t)\right]=\ ф(s)\ \ \ \ then\ \ L\left[\frac{1}{t}f(t)\right]\ =\ \int_s^{\infty{}}ф\left(s\right)ds \\[2ex] \begin{align*} \displaystyle \therefore{}\ L\left[\frac{\sin {2t+\sin {3t}}}{t}\right]&= L\left[\frac{\sin {2t}}{t}\right]\ +\ L\left[\frac{\sin {3t}}{t}\right] \\[2ex] \displaystyle &=\int_s^{\infty{}}\frac{2}{{s^2+2}^2}ds+\ \int_s^{\infty{}}\frac{3}{{s^2+3}^2}ds\ \\[2ex] \displaystyle &=\ {\left[{tan}^{-1}\frac{s}{2}\right]}_s^{\infty{}}\ +\ \ {\left[{tan}^{-1}\frac{s}{3}\right]}_s^{\infty{}} \\[2ex] \displaystyle &=\ \frac{\pi{}}{2}-\ {tan}^{-1}\frac{s}{2}+\frac{\pi{}}{2}-\ {tan}^{-1}\frac{s}{3}\ \ \\[2ex] \displaystyle &=\ \pi{}\ -\ \left({tan}^{-1}\frac{s}{2}+\ {tan}^{-1}\frac{s}{3}\right) \\[2ex] \end{align*} $

$\begin{align*} \displaystyle \therefore{}\ \int_0^{\infty{}}e^{-st}\frac{\sin {2t+\sin {3t}}}{t}dt\ &=\pi{}\ -\ \left({tan}^{-1}\frac{s}{2}+\ {tan}^{-1}\frac{s}{3}\right) \\[2ex] Putting s = 1,&\\[3ex] \displaystyle \int_0^{\infty{}}e^{-t}\frac{\sin {2t+\sin {3t}}}{t}dt&=\ \pi{}\ -\ \left({tan}^{-1}\frac{1}{2}+\ {tan}^{-1}\frac{1}{3}\right) \\[2ex] \displaystyle & =\ \pi{}\ -\ \left(\frac{\pi{}}{4}\right) \\[2ex] \displaystyle &=\ \frac{3\pi{}}{4} \\[2ex] \displaystyle \ \therefore{}\int_0^{\infty{}}\frac{\sin {2t+\sin {3t}}}{te^t}dt&=\ \frac{3\pi{}}{4} \\[2ex] \end{align*}$