Question

Integral og dx/2x-x^2?

Answer 1

`int1/(2x-x^2)dx = "artanh"(x-1) +"c"`

Explanation:

We want to find `1/(2x-x^2)dx`

First complete the square

`int1/(2x-x^2)dx = int1/(-(x^2-2x))dx=int1/-((x-1)^2-1)dx=int1/(1-(x-1)^2)dx`

Now substitute `vartheta=x-1` and `dvartheta = dx`

`int1/(1-(1-x^2))dx=int1/(1-vartheta^2)dvartheta`

This integral can now be found as it's a standard integral

`1/(1-vartheta^2)dvartheta = "artanh"vartheta+"c"`

Now back-substitute for `x`

`"artanh"vartheta+"c"="artanh"(1-x)+"c"`