Calculus of a Single Variable, Chapter 5, 5.7, Section 5.7, Problem 33

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