Answer
$y=$$\frac{1}{2}$$\ln(1+$$x^{2}$)$+C$
Work Step by Step
$\frac{dy}{dx}$$=$$\frac{x}{1+x^{2}}$
$y=$$\int$$\frac{x}{1+x^{2}}$$dx$
if $u=1+$$x^{2}$, $\frac{1}{2x}$$du$$=$$dx$
$y=$$\int$$\frac{x}{u}$$\times$$\frac{1}{2x}$$du$
$y=$$\int$$\frac{1}{2u}$$du$
$y=$$\frac{1}{2}$$\ln(u)$$+C$
$y=$$\frac{1}{2}$$\ln(1+x^{2})$$+C$