How do you find the indefinite integral of #int -(5root4(x))/2dx#?
1 Answer
Jan 24, 2017
Explanation:
We will use the rules:
#intaf(x)dx=aintf(x)dx# #intx^ndx=x^(n+1)/(n+1)+C#
We see that:
#int-(5root4x)/2dx=-5/2intx^(1/4)dx=-5/2(x^(5/4)/(5/4))+C#
Simplifying completely gives:
#=-2x^(5/4)+C#
