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

1 Answer
Feb 3, 2017

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