What is the antiderivative of #int (1/y^3 - 2/sqrty )dy#?

1 Answer
Feb 19, 2016

# -1/(2y^2)-4sqrt(y)+C#

Explanation:

We can re write this as:

#inty^-3-2y^(-1/2)dy#

Now we can integrate by increasing the power by one and dividing by the new power as usual:

#=y^-2/-2-4y^(1/2)+C#

Re writing this back into a more familiar form gives us:

#= -1/(2y^2)-4sqrt(y)+C#