Question

How do you integrate `int (2x-2)/(x^2 - 6x +10)^(1/2)dx` using partial fractions?

Answer 1

`[2sqrt(x^2-6x+10)]+[4arcsinh(x-3)]+C`

Explanation:

`int (2x-2)/sqrt(x^2 - 6x +10)dx`

`int (2x-6+4)/sqrt(x^2 - 6x +10)dx`

`int (2x-6)/sqrt(x^2 - 6x +10)dx+4int1/sqrt(x^2-6x+10)dx`

For the first integral let's `u = x^2-6x+10`

`du = 2x-6 dx`

we have `int1/sqrt(u)du`

which is `[2sqrt(u)]`

complete the square for the second

`x^2-6x+10 = x^2-6x+10 + 9 - 9 = (x-3)^2+1`

`[2sqrt(x^2-6x+10)]+4int1/sqrt((x-3)^2+1)dx`

let's `t = x-3`

`dt = dx`

`[2sqrt(x^2-6x+10)]+4int1/sqrt(t^2+1)dx`

`[2sqrt(x^2-6x+10)]+[4arcsinh(x-3)]+C`