The idea of this homework is to identify some functions $F$ and $G$ such that the RHS is a solution. Note that the derivative of the RHS is
$$
\tfrac32\sqrt{x}(F(x)\log(x)+G(x))+x\sqrt{x}(x^{-1}F(x)+F'(x)\log(x)+G'(x)),
$$
Introduce the function
$$
S(x)=\frac{\sin(x)}x=\sum\limits_{n\geqslant0}s_nx^{2n},\qquad s_n=\frac{(-1)^n}{(2n+1)!}.
$$
Then the derivative of the RHS minus the LHS is
$$
\sqrt{x}\left(\tfrac32G(x)+F(x)+xG'(x)\right)+\sqrt{x}\log(x)\left(\tfrac32F(x)+xF'(x)-S(x)\right),$$ hence the RHS is a solution as soon as
$$
\tfrac32G(x)+F(x)+xG'(x)=0,\qquad
\tfrac32F(x)+xF'(x)=S(x).$$
For the next step, it is probably best to use series expansions.
Assume that
$$
F(x)=\sum\limits_{n\geqslant0}a_nx^{2n},\quad G(x)=\sum\limits_{n\geqslant0}b_nx^{2n}.
$$
The two first order differential equations above yield, for every $n\geqslant0$,
$$
\tfrac32b_n+a_n+2nb_n=0,\qquad\tfrac32a_n+2na_n=s_n,
$$
that is,
$$
a_n=\frac{s_n}{2n+\frac32},\qquad b_n=\frac{-s_n}{(2n+\frac32)^2}.
$$
This determines entirely some functions $F$ and $G$, written as series of infinite radius of convergence, such that the derivative of the RHS is the function in the integral. As a consequence, this determines completely the desired primitives on the domain $x\gt0$.
One can write the functions $F$ and $G$, hence the primitives themselves, in several equivalent ways. The one I prefer is the one above. But if one is fond of generalized hypergeometric functions, one can express $F$ and $G$ with these. For example,
$F(x)=u\left(-\frac{x^2}4\right)$, with
$$
u(z)=\sum\limits_{n\geqslant0}u_nz^n,\qquad u_n=\frac{4^n}{(2n+1)!(2n+\frac32)}.
$$
To write $u_n$ as a ratio of Pochhammer symbols, the procedure is standard. Start from
$$
\frac{u_{n+1}}{u_n}=\frac{4(2n+\frac32)}{(2n+3)(2n+2)(2n+\frac52)}=\frac{n+\frac34}{(n+\frac32)(n+\frac54)}\frac1{n+1}.
$$
Since $u_0=\frac23$, $(a)_0=1$ and $(a)_{n+1}=(a)_n(a+n)$ for every $a$ and every $n\geqslant0$, this yields
$$
u_n=\frac{\frac23(\frac34)_n}{(\frac32)_n(\frac54)_n}\frac1{n!}.
$$
Finally,
$$
F(x)=\frac23\,{}_1\!F_2\left(\frac34;\frac32,\frac54;-\frac{x^2}4\right).
$$
Likewise, one can express $G$ in terms of a hypergeometric function of type ${}_2\!F_3$, an exercise we will leave to the reader.