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?
 A: Hint: the formula $$\int x^c\, dx = \frac{x^{c+1}}{c+1}  + C$$
applies to all real numbers $c \neq -1$, not just integers.
A: In derivative we subtract 1.
In integration we add 1.
Your solution is fine. Except last step.
$$=\int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2}) dx $$
$$=\left[2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2}\right]_0^1$$
$$=\left[2\cdot1^{1/2} + 2\cdot1^{3/2} + \frac{10}{7}\cdot1^{7/2}\right]-\left[2\cdot0^{1/2} + 2\cdot0^{3/2} + \frac{10}{7}\cdot0^{7/2}\right]$$
Hope now you can proceed.
A: It looks like you've already done the integration correctly, but forgot to take off the integration sign. The last line should be
$$\left[2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2} \right]_0^1$$
which you can evaluate by substituting $x=1$ and $x=0$, and subtracting.
You asked whether you should substitute $x=1$: this will happen to give you the right answer, but only coincidentally, because the expression becomes $0$ when $x=0$.
A: Putting $t=\sqrt{x}$ you have $dt=\frac{1}{2\sqrt{x}}dx$ and the limits stay the same.
$$\int_0^12(1+3t^2+5t^6)$$
A: 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 $F(1)-F(0)$.
A: We put $t=\sqrt{x}$, so $dt=\frac{1}{2\sqrt{x}}dx$.
$$\int_0^12(1+3t^2+5t^6)$$
$$=2\left[t+t^3+\frac{5t^7}{7}\right]^1_0$$
$$=2\left[\sqrt{x}+\sqrt{x}^3+\frac{5\sqrt{x}^7}{7}\right]^1_0$$
$$=\frac{38}{7}$$
