Evaluating: $\int 3x\sin\left(\frac x4\right) \, dx$. 
$\displaystyle\color{darkblue}{3\int x\sin\left(\dfrac x4\right)\,\mathrm dx}$
$$\begin{align}
\dfrac{-4x}{x}\cos\dfrac x4 \,\,\boldsymbol\Rightarrow\,\, & -4\cos\left(\dfrac x4\right)-\int \dfrac{-4}{x}\cos\left(\dfrac x4 \right)\,\mathrm dx\\\,\\
& 3\left(-4\cos\left(\dfrac x4\right)+\int\dfrac4x\cos\left(\dfrac x4\right)\,\mathrm dx\right)\\\,\\
&\int\dfrac{4\cos\left(\frac x4\right)}{x}\mathrm dx
\end{align}$$ $\displaystyle \color{darkblue}{uv-\int v\dfrac{\mathrm du}{\mathrm dx}\,\mathrm dx }$ 
$\displaystyle\boxed{\displaystyle\,\,-4\cos\left(\dfrac x4\right)+4\int\dfrac{\cos x/4}{x}\,\mathrm dx\,\,}$
$\displaystyle 3\left[-4\cos\left( \dfrac x4\right) +4\left(\cos\left( \dfrac x4\right)\ln(x)\right)\right]$
$\displaystyle -12\cos (x/4) + 12\cos (x/4) \ln(x) \rightarrow \text{ wrong.}$

$\displaystyle\color{darkblue}{\int 3x\sin\left(\dfrac x4\right)\,\mathrm dx}$
  $\quad\quad\quad\quad\displaystyle\int\dfrac{\cos(x/4)}{x}\rightarrow\dfrac1x\int\cos(x/4)\mathrm dx$
$\displaystyle3\left[-4\cos(x/4)+\dfrac4x\sin(x/4)\right]$
$\displaystyle -12\cos(x/4)+\dfrac{48}{x}\sin(x/4)$

In the text above is my work done to solve the following question: 


Find the indefinite integral of: $\left[3x\sin\left(\dfrac x4\right)\right]$ 


The bordered area is the furthest I got (there should be a 3 at the front to multiply the whole equation but I usually remember to add that at the end) The part where I wrote "wrong" is what I thought the answer was, I assumed it would be such. What I'm having troubles with is integrating the $\frac{\cos(x/4)}{x}$. Would I need to make it $\frac{\cos(x/4)}{x}$ and then integrate by parts to get that integral? Thanks in advance, I hope I made some sense in what I'm trying to achieve.  I guess what I'm looking for, is a way to integrate $$ \frac{\cos(x/4)}{x}$$
 A: I think you have made a mistake in the application of integration by parts. Taking $u$ as $x$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ as $\sin \frac{x}{4}$, we should get 
\begin{align*} 3 \int {x} \sin \frac{x}{4} \mathrm{d}x &= 3 ( -4x \cos \frac{x}{4}- \int -4 \cos \frac{x}{4} \mathrm{d}x) \\ &= -12x \cos \frac{x}{4} + 48 \sin \frac{x}{4} + C.
\end{align*}
A: Note that $$\left(-\frac{x}{a}\cos (ax)+\frac{1}{a^2}\sin (ax) \right)'  = x\sin( ax)$$
that is 
$$\color{red}{\int x\sin( ax)dx= \left(-\frac{x}{a}\cos (ax)+\frac{1}{a^2}\sin (ax) \right)+c } $$
Then now you can take $a=\frac14$
A: $$I=\int3x \sin(\frac{x}{4})dx\\
u=\frac{x}{4}\Rightarrow du=\frac{dx}{4}\\
I=48\int u\sin(u)du=-48\int ud(\cos(u))=-48u\cos(u)+48\sin(u)+c\\
=48\sin(\frac{x}{4})-12x\cos(\frac{x}{4})+c\\
$$
A: I know this wasn't exactly asked for - but this is a "nice to have" add on in my opinion.
Another method which I've found rather interesting: differentiation under the integral sign (popularised by Feynman).
Consider the integral: $\displaystyle I = \int x\sin{kx} dx$, where $k$ is a constant.
Note that $\displaystyle I = \frac{\partial}{\partial k}\int (-\cos kx)dx = \frac{\partial}{\partial k}(-\frac{1}{k}\sin kx + c) = \frac 1{k^2}\sin kx - \frac xk\cos kx$
Now to get the desired integral, substitute $\displaystyle k= \frac 14$ and remember to multiply by $3$ to get the desired integral (and don't forget the constant):
$\displaystyle \int 3x \sin{\frac x4}dx = 3(\frac 1{(\frac 14)^2}\sin (\frac 14)x - \frac x{\frac 14}\cos (\frac 14)x) + c = 48\sin{\frac x4}- 12\cos{\frac x4} + c$
To address your other sub-question, note that functions like $\frac{\cos x}{x}$ and $\frac{\sin x}{x}$ (give or take a constant multiplier in the argument) do not have elementary anti-derivatives. That means that you need to use special functions - either you have to define a special function to cover this case or compute it in terms of other special functions. Since defining these special functions often involves integrals anyway, the whole process is rather circular.
Long story short, there is no elementary way to compute the indefinite integral you're now asking about. But note that the definite integral is solvable for particular bounds, but that's another story.
