Answer
$ y = \sqrt {x^2 + 2ln|x| + C}$ or $-\sqrt {x^2 + 2ln|x| + C }$
Work Step by Step
$xyy' = x^2 + 1$ can be rewritten as:
$xy (dy/dx) = x^2 + 1$
Now let's separate y on one side and x on the other side:
$y dy = ((x^2 + 1)/x )dx$
Now let's integrate both sides:
$\int y dy = \int((x^2 + 1)/x )dx$
$y^2/2 = \int(x+ 1/x )dx$
$y^2/2 = x^2/2 + ln|x| + C_1 $
$y^2 = x^2 + 2ln|x| + 2C_1 $
Let's replace $2C_1 = C$
$ y = \sqrt {x^2 + 2ln|x| + C }$ or $-\sqrt {x^2 + 2ln|x| + C} $
