Given #int(2x-5)/(x^2+2x+2)dx#
You can manipulate the expression* into
#color(red)(int((2x+2)/(x^2+2x+2))dx) - int(7/(x^2+2x+2))dx#
For the first integral,
let #color(red)(u= x^2 +2x+2) ; du = (2x+2) dx #
#int(du)/u = ln|u| +C#
#int((2x+2)/(x^2+2x+2))dx =color(red)( ln(x^2 + 2x+2) +C#
For the second integral
# - int(7/(x^2+2x+2))dx# , let's complete the square for the denominator
#color(blue)(-int7/((x^2+2x+1)+2-1)dx#
#=> int( -7)/((x+1)^2 + 1)dx#
Let #v= (x+1) ; dv= 1dx " " ; a^2 = 1 ; a= 1#
#int1/(u^2+a^2)du = arctan (u/a) +C#
#=>-7int1/((x+1)^2 + 1)dx =>color(blue)( -7 arctan (x+1) + C#
When we put it together we have
#int(2x-5)/(x^2+2x+2)dx = color(red)( ln(x^2 + 2x+2) color(blue)( -7 arctan (x+1) + C#
*Why? Because if we differentiate the denominator, we ALMOST have the derivative on the numerator...
(If you don't like this method you can also use partial fraction decomposition but it will take longer)
