Answer
Use integration by parts with $u=x$ and $dv=sin(3x+4)dx$
Work Step by Step
Given $$ \int x \sin (3 x+4) d x$$
Let
\begin{align*}
u&= x\ \ \ \ \ \ dv= \sin (3x+4)\\
du&= dx\ \ \ \ \ v=\frac{-1}{3}\cos(3x+4)
\end{align*}
Then
\begin{align*}
\int x \sin (3 x+4) d x&=\frac{1}{3} x \cos (3 x+4)+\frac{1}{3} \int \cos (3 x+4) d x
\end{align*}
