Question

How do you integrate `int (2x-2)/ ((8 + 2x - x^2)^(1/2))dx` using partial fractions?

Answer 1

The answer is `=-2sqrt(8+2x-x^2)+C`

Explanation:

You can integrate only by substitution.
Let `u=8+2x-x^2` `=>` `du=(2-2x)dx=-(2x-2)dx`
`int((2x-2)dx)/(sqrt(8+2x-x^2))=-int(du)/sqrtu=-u^(-1/2+1)/(-1/2+1)`
`=-2sqrtu=-2sqrt(8+2x-x^2)+C`