Find the integral of 4sqrt(x) - 6/x^3.

The easiest way to do this is to break the integral up into it's separate parts. We have:integral of 4sqrt(x)andintegral of -6/x^3
Both of which have constants that can be taken out, i.e. 4 and 6, respectively.So all we need to compute is the integral of sqrt(x), and the integral of 1/x^3. These can be rewritten as x^(1/2), and x^-3, respectively. With the terms in this form it's easier to see that we can use the properties of polynomials to do these computations. sqrt(x) = x^(1/2) integrates to (x^(3/2))/(3/2) = 2/3 * x^(3/2), and 1/x^3 = x^-3 integrates to x^-2/-2 = -1/2x^2.
We can substitute these values back into our integral to see that 4sqrt(x) integrates to 4 * 2/3 * x^(3/2) = 8/3 * x^(3/2), and 6/x^3 integrates to -6 * -1/2x^2 = 3/x^2. The final step is to add the integrating constant, c, as this is an indefinite integral. Hence the final answer is 8/3 * x^(3/2) + 3/x^2 + c.