Question

How do you integrate `h(t)=root3t(t^2+4)` using the product rule?

Answer 1

The integral is `(3/10)t^(10/3)+(16/3)t^(4/3)+C` which can be simplified to: `(t^(2/3)/10)(3t^5+16t^2)+C`

Explanation:

The most important thing to understand is that there is no product rule or quotient rule for Integration. This is a mistake many students make. Sometimes, you can use u-substitution to deal with a problem like this, but first you should try something simpler. Let's use algebra to simplify it. Remember that `root(3)t=t^(1/3)`.

So, this is:
`int(t^(1/3)(t^2+4))`

`int(t^(7/3)+4t^(1/3))` We now use the power rule:

`t^(10/3)/(10/3)+(4t^(4/3)/(4/3))+C` `=`

`(3/10)t^(10/3)+(16/3)t^(4/3)+C`

If you enjoy simplifying, you could do this:

`(9/30)t^(10/3)+(48/30)t^(4/3)+C`=

`(3/30)t^(2/3)*(3t^5+16t^2)+C` = `(t^(2/3)/10)(3t^5+16t^2)+C`