Integrate $\frac{x^2-4x+10}{x^2\sqrt x}$ 
Find the indefinate integral with respect to $x$ of $$\frac{x^2-4x+10}{x^2\sqrt x}$$

For this problem I first made each individual number in the numerator separate from each other for easier integration and then simplified
$$=\int \left(\frac{x^2}{x^2\sqrt x}-\frac{4x}{x^2\sqrt x}+\frac{10}{x^2\sqrt x}\right)dx$$
$$=\int \left(x^2x^{-\frac{3}{2}}-4x^1x^{-\frac{3}{2}}+10x^{-\frac{3}{2}}\right)dx$$
$$=\int \left(x^{\frac{1}{2}}-4x^{-\frac{1}{2}}+10x^{-\frac{3}{2}}\right)dx$$
I then integrated the expression to get
$$\frac{2x^{\frac{3}{2}}}{3}-8x^{\frac{1}{2}}-20x^{-\frac{1}{2}}$$
This is, however, wrong. Any ideas as to why?
Thanks in advance $:)$

You could have noticed that the $x^2$ simplify in the first term,
$$\frac{x^2+\cdots}{x^2\sqrt x}=\frac1{\sqrt x}+\cdots=x^{-1/2}+\cdots$$
So with a little more care,
$$\int(x^{-1/2}-4x^{-3/2}+10x^{-5/2})\,dx=\frac21x^{1/2}-\frac21(-4)x^{-1/2}-\frac2310x^{-3/2}.$$
