Computing $ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$ $$ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$$
My idea for this was to break each numerator into its own fraction as follows
$$ \int_0^1 \left(\frac{1}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{5x^3}{\sqrt{x}}\right)dx$$
$$ \int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2})\ dx $$
$$ \int_0^1 2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2} $$
Not really sure where to go from there. Should I sub 1 in for the x values and let that be the answer?

Your work is correct, but for the last step write:
$$
\left[ 2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}\right]_0^1
$$
because you have just done the integration as antiderivative: $F(x)=2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}$ and you have only to evaluate the primitive at the two limits of integration, so that your definite integrale is done by $F(1)-F(0)$.
