Asked by kmac
integral (x^2+x-1)/(x+1) dx can you please explain maybe not step by step but not just the answer please thanks
Answers
Answered by
Reiny
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
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