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

1 Answer
Feb 3, 2017

Answer:

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#