## Calculus: Early Transcendentals 8th Edition

$y = \sqrt {x^2 + 2ln|x| + C}$ or $-\sqrt {x^2 + 2ln|x| + C }$
$xyy' = x^2 + 1$ 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 $-\sqrt {x^2 + 2ln|x| + C}$