`int_0^2 dx/(x^2-2x+2) ` Find or evaluate the integral by completing the square

To be able to evaluate the given integral:` int_0^2 (dx)/(x^2-2x+2)` , we

complete the square of the expression:`x^2-2x+2` .

To complete the square, we add and subtract `(-b/(2a))^2` .

 The `x^2-2x+2` resembles the `ax^2+bx+c` where:

 `a=1` , `b =-2 ` and `c=2 ` .

Then,

`(-b/(2a))^2 =(-(-2)/(2(1)))^2`

             ` =(2/2)^2`

             ` = (1)^2`

             ` =1`

Add and subtract 1 :

`x^2-2x+2 +1-1`

Rearrange as: `(x^2-2x +1) +2-1 = (x-1)^2+1`

Plug-in` x^2-2x+2 = (x-1)^2+1` in the given integral:`int_0^2 (dx)/(x^2-2x+2)` .

`int_0^2 (dx)/(x^2-2x+2) =int_0^2 (dx)/((x-1)^2+1)`

This resembles the basic integral formula for inverse tangent function:

`int (du)/(u^2+a^2) =1/a *arctan(u/a)+C`

Then  indefinite integral F(x)+C,

`int (dx)/((x-1)^2+1^2) =1/1 *arctan((x-1)/1)+C`

                                `=arctan(x-1)+C` 

For the definite integral, we apply: `F(x)|_a^b= F(b)-F(a)` .

`arctan(x-1)|_0^2 =arctan(2-1) -arctan(0-1)`

                    ` =arctan(1) -arctan(-1)`

                    ` =pi/4 -(-pi/4)`

                    ` =pi/4 +pi/4`

                    ` =(2pi)/4`

                    ` =pi/2`