Answer
$y = c(x^2+1)$
Work Step by Step
Separating variables, we have
$\frac{y'}{y} = \frac{2x}{1+x^2}\implies \frac{dy}{y} = \frac{2x}{x^2+1}dx$
Integrating both sides yields
$\ln{|y|} = \int{\frac{2x}{x^2+1}dx}$
Using $u$-substitution with $u = x^2+1, du=2x dx,$ we have
$\ln{|y|} = \int{\frac{du}{u}}=\ln{|u|}+C=\ln{|x^2+1|}+C$
Note that for any real number $C$, there is a positive $k$ such that $C = \ln{k}$, so
$\ln{|y|} = \ln{|k(x^2 + 1)|}$
Dropping the $\ln$ on both sides yields
$|y| = |k(x^2 + 1)|$
From which it can be deduced that
$y = \pm k(x^2+1)\implies y=c(x^2+1)$
For $c$ any real number (positive or negative).
