Solution to the $\displaystyle \phi'' + 4\phi' + 3\phi = x \cos {3x}. $ I came across the following differential equation while practicing  $ \displaystyle \phi'' + 4\phi' + 3\phi  = x \cos {3x} .$
Now, I used VARIATION OF PARAMETERS to solve it. I found general solution the equation $ \displaystyle \phi'' + 4\phi' + 3\phi  = 0 $ to be $\displaystyle C_1 e^{-3x} + C_2 e^{-x}.$
Now we need to find two functions $\lambda \ \text{and} \ \nu $ such that $\displaystyle \lambda'e^{-3x} + \nu'e^{-x} = 0 $ and $\displaystyle -3\lambda'e^{-3x} - \nu'e^{-x} = x \cos {3x} .$
Then the general solution would be : $\displaystyle \lambda e^{-3x} + \nu e^{-x} .$
On solving I got : $$\displaystyle \lambda = -\frac {1}{2} \int x \cdot \cos{3x} \cdot e^{3x} \mathrm {d}x \\  \nu = \frac {1}{2} \int x \cdot \cos{3x} \cdot e^{x} \mathrm {d}x  $$
These integrals took me a long time to solve. I used the traditional DI method to solve for it.
Finally getting the answer as : $$ \displaystyle C_1 e^{-3x} + C_2 e^{-x} + \frac {x}{30} ( 2 \sin {3x} - \cos {3x}  ) + \frac {47}{1800}\cos {3x}  +\frac {19}{200}\sin {3x} .  $$
I want my answer to get verified and am open to some good way to the integral quickly.
Thank you in advance.

You can also apply the method of undetermined coefficients to find the particular solution. The solution to the complementary differential equation   $\phi'' + 4\phi' + 3\phi=0$ is
$$\phi_c(x) =c_1e^{-3x}+c_2e^{-x}.$$
Thus to find the particular solution to $\phi'' + 4\phi' + 3\phi = x \cos {3x}$ we try a particular solution of the form
$$\phi_p(x)=(Ax+B)\cos(3x)+(Cx+D)\sin(3x).$$
Hence
$$\phi_p'(x)=(A + 3Cx + 3D)\cos(3 x)+(-3 A x - 3 B + C) \sin(3 x), $$
$$\phi_p''(x)= (-9 A x - 9 B + 6 C)\cos(3 x) +  (-6 A -9Cx-9D)\sin(3 x).$$
Plugging these into the differential equation gives
\begin{equation}
\begin{split}
(-6A+12C)x\cos(3x) +(-6C-12A)x\sin(3x)+(4A-6B+6C+12D)\cos(3x)\\
+(-6A-12B+4C-6D)\sin(3x)=x\cos(3x).
\end{split}
\end{equation}
Therefore by relating coefficients
$$-6C-12A =0 \implies C=-2A,$$
$$-6A+12C=1 \implies \boxed{A=-\frac{1}{30}}, \quad \boxed{C=\frac{1}{15}},$$
$$4A-6B+6C+12D=0 \implies -6B+12D = -\frac{4}{15}\implies D = \frac{B}{2} -\frac{1}{45},$$
$$-6A-12B+4C-6D=0 \implies -15B+\frac{2}{15}=-\frac{7}{15}\implies \boxed{B=\frac{1}{25}}, \quad \boxed{D= -\frac{1}{450}}.$$
So the particular solution becomes
$$\phi_p(x)=\left(-\frac{1}{30}x+\frac{1}{25}\right)\cos(3x)+\left(\frac{1}{15}x-\frac{1}{450}\right)\sin(3x).$$
