integral (x^2+x-1)/(x+1) dx can you please explain maybe not step by step but not just the answer please thanks
first do a long division
(x^+x-1)/(x+1) = x - 1/( +1)
so ∫(x^+x-1)/(x+1) dx
= ∫(x - 1/(x +1)) dx
= (1/2)x^2 - ln(x+1) + c
(x^+x-1)/(x+1) = x - 1/( +1)
so ∫(x^+x-1)/(x+1) dx
= ∫(x - 1/(x +1)) dx
= (1/2)x^2 - ln(x+1) + c