solve indefinite integral I have this indefinite integral $\int 3 \sqrt{x}\,dx$ to solve.
My attempt:
$$\int 3 \sqrt{x}\,dx = 3 \cdot \frac {x^{\frac {1}{2} + \frac {2}{2}}}{\frac {1}{2} + \frac {2}{2}}$$
$$\int 3 \sqrt{x}\,dx = 3 \frac{x^{\frac {3}{2}}}{\frac {3}{2}} = \frac{2}{3} \cdot \frac{9}{3} x^{\frac {3}{2}}$$
$$\int 3 \sqrt{x}\,dx = \frac{18}{3} x^{\frac{3}{2}} = 6 x^{\frac{3}{2}}$$
But according to wolframalpha the answer should be $2 x^{\frac {3}{2}}$
Where did I make a error in my calculation?
Thanks!

$\frac{2}{3} \cdot \frac{9}{3}=2$. not $6$
