Calculus, 10th Edition (Anton)

$=\int{x^{2}dx}=\frac{1}{3}x^{3}+C$
$2ln(x)$ can be written as $ln(x^{2})$ by bringing the constant multiplier into the logarithm. This leaves $e^{ln(x^{2})}$, which equals simply $x^{2}$. This leaves $$\int{x^{2}dx}=\boxed{\frac{1}{3}x^{3}+C}$$