Answer
$$
x \ln x =y(1+\sqrt {3+y^{2}})y^{\prime }, \quad y(1)=1 .
$$
The solution of the given differential equation is
$$
\frac{1}{2}x^{2}\ln x-\frac{1}{4} x ^{2}+\frac{41}{12}=\frac{1}{2} y^{2} +\frac{1}{3 }(3+y^{2})^{\frac{3}{2}}.
$$
Work Step by Step
$$
x \ln x =y(1+\sqrt {3+y^{2}})y^{\prime }, \quad y(1)=1 .
$$
We write the equation in terms of differentials:
$$
( x \ln x) dx =y(1+\sqrt {3+y^{2}})dy
$$
and integrate both sides:
$$
\int ( x \ln x) dx =\int(y+y\sqrt {3+y^{2}})dy
$$
it can be written in the form
$$
I_{1} =\frac{1}{2} y^{2} + I_{2} \quad (1)
$$
First, we can find the following integral:
$$
I_{1}= \int ( x \ln x) dx
$$
use integration by parts with
$$
\quad\quad\quad \left[\begin{array}{c}{u=\ln x, \quad\quad dv= x dx } \\ {d u= \frac{dx}{x} , \quad\quad v= \frac{1}{2}x^{2} }\end{array}\right] , \text { then }\\
$$
$ \Rightarrow$
$$
I_{1}= \frac{1}{2}x^{2}\ln x-\frac{1}{2}\int x dx=\frac{1}{2}x^{2}\ln x-\frac{1}{4} x ^{2}+C_{1}, \quad (2)
$$
where $C_{1} $ is an arbitrary constant.
and
$$
I_{2}=\int y\sqrt {3+y^{2}}dy
$$
put $w=3+y^{2}$ then $dw =2ydy$ so,
$$
I_{2}=\frac{1}{2}\int \sqrt {w}dw=\frac{1}{3 }(3+y^{2})^{\frac{3}{2}}+C_{2}, \quad (3)
$$
where $C_{2}$ is an arbitrary constant.
substituting from Eq. (2)and Eq. (3) in Eq.(1) we have:
$$
\frac{1}{2}x^{2}\ln x-\frac{1}{4} x ^{2}+C=\frac{1}{2} y^{2} +\frac{1}{3 }(3+y^{2})^{\frac{3}{2}}
$$
where C is an arbitrary constant.
Now, to satisfy the condition $ y(1)=1, $we have
$$
0-\frac{1}{4}+C=\frac{1}{2} (1)^{2} +\frac{1}{3 }(3+(1)^{2})^{\frac{3}{2}}
$$
$ \Rightarrow$
$$
C=\frac{1}{2} +\frac{8}{3 }+\frac{1}{4}=\frac{41}{12}.
$$
So, the solution of the given differential equation is
$$
\frac{1}{2}x^{2}\ln x-\frac{1}{4} x ^{2}+\frac{41}{12}=\frac{1}{2} y^{2} +\frac{1}{3 }(3+y^{2})^{\frac{3}{2}}
$$
