Answer
$\displaystyle \frac{x^2}{2}\ln(x) -\frac{x^2}{4} + \frac{41}{12} = \frac{y^2}{2} + \frac{(3+y^2)^{3/2}}{3}$
Work Step by Step
$\displaystyle x\ln(x) = y(1+\sqrt{3+y^2})y'$
Separation of Variables
$\displaystyle x\ln(x) = (y+y\sqrt{3+y^2})\frac{dy}{dx}$
$\displaystyle \int[x\ln(x)]dx = \int(y+y\sqrt{3+y^2})dy$
Integrate each side. The left side uses integration by parts, and the right side uses $u$-substitution:
Integration by parts formula: $\displaystyle \int (f\cdot g')dx = f\cdot g-\int (f'\cdot g)dx$ For the left side, let $f = \ln(x)$ and $g' = x$. For the right side let $u = 3 + y^2$ so that $du = 2y$
$\displaystyle \ln(x) \cdot \frac{x^2}{2} -\int \frac{x^2}{2}\cdot \frac{1}{x}dx = \int y\cdot dy+\int y\sqrt{3+y^2})dy$
$\displaystyle \frac{x^2}{2}\ln(x) -\int \frac{x}{2}dx = \frac{y^2}{2} +\frac{1}{2}\int u^{1/2}du$
$\displaystyle \frac{x^2}{2}\ln(x) -\frac{x^2}{4} + C = \frac{y^2}{2} +\frac{1}{2}(\frac{u^{3/2}}{3/2})$
$\displaystyle \frac{x^2}{2}\ln(x) -\frac{x^2}{4} + C = \frac{y^2}{2} +\frac{u^{3/2}}{3}$
$\displaystyle \frac{x^2}{2}\ln(x) -\frac{x^2}{4} + C = \frac{y^2}{2} + \frac{(3+y^2)^{3/2}}{3}$
Plug in the initial condition to solve for $C$
$\displaystyle \frac{(1)^2}{2}\ln(1) -\frac{(1)^2}{4} + C = \frac{(1)^2}{2} + \frac{(3+(1)^2)^{3/2}}{3}$
$\displaystyle 0 - \frac{1}{4} + C = \frac{1}{2} + \frac{(3+1)^{3/2}}{3}$
$\displaystyle C = \frac{1}{2} + \frac{(4)^{3/2}}{3} + \frac{1}{4}$
$\displaystyle C = \frac{3}{4} + \frac{(2)^{3}}{3}$
$\displaystyle C = \frac{3}{4} + \frac{8}{3} = \frac{9}{12} + \frac{32}{12} = \frac{41}{12}$
Plug $C$ back into the equation for the final answer
$\displaystyle \frac{x^2}{2}\ln(x) -\frac{x^2}{4} + \frac{41}{12} = \frac{y^2}{2} + \frac{(3+y^2)^{3/2}}{3}$
