How to integrate (x^2+4x-9/+2)?
Well, we could make the long division but we see by looking at the two first terms of the numerator that we can write (x^2+4x-9)/(x+2) = (x^2 + 2*2x + 2^2 - 2^2- 9)/(x+2) = ( (x+2)^2 - 13)/(x+2) <--- so we obtain the result = x+2 - 13/(x+2) therefore : ∫ (x^2+4x-9)/(x+2) dx = ∫ [ x+2 - 13/(x+2) ] dx = x^2 /2 + 2x - 13 ln( |x+2| ) + C qed hope it' ll help !! PS : welcome to the forum !! if you want good answers, don't forget to give BAs too !! ;-)
Presentation looks dodgy !!!!