I need a lot of help understanding this as I am completely lost at what to do
Evaluate to 4 significant digits: $\displaystyle \int_{-1} ^5\;\frac{5x^2}{\sqrt{x^3+3}}\;dx$
Ok, first step is to do a little rearranging of constants inside and outside the integral. Since you can pull out a constant, begin with
$\displaystyle \frac{10}{3}\int_{-1} ^5\;\frac{3x^2}{2\sqrt{x^3+3}}\;dx$.
Then you have inside the integral the derivative of $\displaystyle \sqrt{x^3 + 3}$, so you can evaluate to $\displaystyle \frac{10}{3}(\sqrt{5^3 + 3}-\sqrt{(-1)^3 + 3})$.
It might help to view this as $\displaystyle \int_{-1} ^5\;{5x^2}(x^3+3)^{\frac{-1}{2}}\;dx$
Solving integrals a lot of times means being aware of the form $\displaystyle \int_{a}^{b}\; u^n du$ where in this case $\displaystyle u = x^3 + 3, n = \frac{-1}{2}, du = 3x^2 dx$
if you can't see what to multiply by to get everything in the proper form, it is ok. just solve for it.
Let $\displaystyle u = x^3 + 3$
$\displaystyle \Rightarrow du = 3x^2~dx$
$\displaystyle \Rightarrow \frac 13 ~du = x^2~dx$ ................(you could also have said $\displaystyle \frac 53~du = 5x^2 ~dx$)
So now you just replace $\displaystyle x^2~dx$ with $\displaystyle \frac 13~du$ and you integral now becomes:
$\displaystyle \frac 53 \int \frac 1{\sqrt{u}}~du = \frac 53 \int u^{- 1/2}~du$
Alternatively, you could let $\displaystyle u^2 = x^3 + 3$ and that would result in the integral $\displaystyle \frac {10}3 \int 1 ~du$