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?

We put $t=\sqrt{x}$, so $dt=\frac{1}{2\sqrt{x}}dx$.
$$\int_0^12(1+3t^2+5t^6)$$
$$=2\left[t+t^3+\frac{5t^7}{7}\right]^1_0$$
$$=2\left[\sqrt{x}+\sqrt{x}^3+\frac{5\sqrt{x}^7}{7}\right]^1_0$$
$$=\frac{38}{7}$$
