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Topic: **Homework notice form****Question:**
What is the antiderivative of: (x+1) / ( (x^2) -x +1 ) ?
Jarrod, your answer is wrong... yo take the numerator as 2x - 2. Derivate your answer and you will notice that is different than the antiderivative i wrote.
Northstar, germano, thank you so much for your answers. If i could choose two best answers i choose yours. I can't choose one of you because the two answers are great. Thank you so much both of you.
By the way, Kate. This antiderivative is not a homework. I'm doing practice exercises and i found some complicated antiderivatives. I have exam next week. I just want to be ready. And i appreciate a lot the people who answer my question.
Well, i think is fair to choose and give points to the best answer. I choose germano because he wrote a quiet explanation for my question. But i really appreciate your answer, Northstar.

June 16, 2019 / By Karolyn

∫ [(x + 1)/(x² - x + 1)] dx = since the denominator doesn't factor (in that it has no real zeros), make the top the derivative of the bottom, dividing and multiplying by 2: (1/2) ∫ [2(x + 1)/(x² - x + 1)] dx = (1/2) ∫ [(2x + 2)/(x² - x + 1)] dx = to complete the derivative on the top, add and subtract - 1: (1/2) ∫ [(2x - 1 + 1 + 2)/(x² - x + 1)] dx = (1/2) ∫ {[(2x - 1) + 3]/(x² - x + 1)} dx = break it up into: (1/2) ∫ (2x - 1) dx/(x² - x + 1) + (1/2) ∫ [3 /(x² - x + 1)] dx = (1/2) ∫ d(x² - x + 1)/(x² - x + 1) + (3/2) ∫ dx /(x² - x + 1) = (1/2) ln(x² - x + 1) + (3/2) ∫ dx /(x² - x + 1) = complete the square by breaking 1 into (1/4) + (3/4): (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{[x² - x + (1/4)] + (3/4)} = (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{[x - (1/2)]² + (3/4)} = factor (3/4) out of the bottom, so as to get it into the form [f(x)]² + 1: (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{(3/4) {(4/3)[x - (1/2)]² + 1}} = include (4/3) into the square: (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{(3/4) { {(2/√3)[x - (1/2)]}² + 1} } expand the base of the square and shift (3/4) to the top (as 4/3): (1/2) ln(x² - x + 1) + (3/2) ∫ (4/3)dx /{[(2/√3)x - (1/√3)]² + 1} = split (4/3) into (2/√3)(2/√3) and pull (2/√3) out, yielding (2/√3)dx, that is d[(2/√3)x - (1/√3)]: (1/2) ln(x² - x + 1) + (3/2) ∫ (2/√3)(2/√3)dx /{[(2/√3)x - (1/√3)]² + 1} = (1/2) ln(x² - x + 1) + (3/2)(2/√3) ∫ (2/√3)dx /{[(2/√3)x - (1/√3)]² + 1} = (1/2) ln(x² - x + 1) + (√3) ∫ d[(2/√3)x - (1/√3)] /{[(2/√3)x - (1/√3)]² + 1} = that is of the form ∫ d[f(x)] /{[f(x)]² + 1} = arctan[f(x)] + C so you conclude with: ∫ [(x + 1)/(x² - x + 1)] dx = (1/2) ln(x² - x + 1) + (√3)arctan[(2/√3)x - (1/√3)] + C I hope it helps..

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∫ [(x + 1)/(x² - x + 1)] dx = since the denominator doesn't factor (in that it has no real zeros), make the top the derivative of the bottom, dividing and multiplying by 2: (1/2) ∫ [2(x + 1)/(x² - x + 1)] dx = (1/2) ∫ [(2x + 2)/(x² - x + 1)] dx = to complete the derivative on the top, add and subtract - 1: (1/2) ∫ [(2x - 1 + 1 + 2)/(x² - x + 1)] dx = (1/2) ∫ {[(2x - 1) + 3]/(x² - x + 1)} dx = break it up into: (1/2) ∫ (2x - 1) dx/(x² - x + 1) + (1/2) ∫ [3 /(x² - x + 1)] dx = (1/2) ∫ d(x² - x + 1)/(x² - x + 1) + (3/2) ∫ dx /(x² - x + 1) = (1/2) ln(x² - x + 1) + (3/2) ∫ dx /(x² - x + 1) = complete the square by breaking 1 into (1/4) + (3/4): (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{[x² - x + (1/4)] + (3/4)} = (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{[x - (1/2)]² + (3/4)} = factor (3/4) out of the bottom, so as to get it into the form [f(x)]² + 1: (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{(3/4) {(4/3)[x - (1/2)]² + 1}} = include (4/3) into the square: (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{(3/4) { {(2/√3)[x - (1/2)]}² + 1} } expand the base of the square and shift (3/4) to the top (as 4/3): (1/2) ln(x² - x + 1) + (3/2) ∫ (4/3)dx /{[(2/√3)x - (1/√3)]² + 1} = split (4/3) into (2/√3)(2/√3) and pull (2/√3) out, yielding (2/√3)dx, that is d[(2/√3)x - (1/√3)]: (1/2) ln(x² - x + 1) + (3/2) ∫ (2/√3)(2/√3)dx /{[(2/√3)x - (1/√3)]² + 1} = (1/2) ln(x² - x + 1) + (3/2)(2/√3) ∫ (2/√3)dx /{[(2/√3)x - (1/√3)]² + 1} = (1/2) ln(x² - x + 1) + (√3) ∫ d[(2/√3)x - (1/√3)] /{[(2/√3)x - (1/√3)]² + 1} = that is of the form ∫ d[f(x)] /{[f(x)]² + 1} = arctan[f(x)] + C so you conclude with: ∫ [(x + 1)/(x² - x + 1)] dx = (1/2) ln(x² - x + 1) + (√3)arctan[(2/√3)x - (1/√3)] + C I hope it helps..

Find the antiderivative. (x + 1)/(x² - x + 1) Let u = x - 1/2 du = dx Then we have: (x + 1) / (x² - x + 1) = (u + 3/2) / (u² + 3/4) Now we can integrate. ∫[(x + 1)/(x² - x + 1)]dx = ∫[(u + 3/2)/(u² + 3/4)]du = (1/2)∫[2u/(u² + 3/4)]du + ∫[(3/2)/(u² + 3/4)]du = (1/2)ln(u² + 3/4) + ∫[(3/2)/(u² + 3/4)]du For the remaining integral let (√3/2)tanθ = u (√3/2)sec²θ dθ = du __________ = (1/2)ln(u² + 3/4) + (3/2)∫{1/[(3/4)tan²θ + 3/4] [(√3/2)sec²θ dθ]} = (1/2)ln(u² + 3/4) + √3∫dθ = (1/2)ln(u² + 3/4) + (√3)θ = (1/2)ln(x² - x + 1) + (√3)arctan[(2/√3)u] + C = (1/2)ln(x² - x + 1) + (√3)arctan[(2/√3)(x - 1/2)] + C = (1/2)ln(x² - x + 1) + (√3)arctan[(2x - 1)/√3] + C

Find the antiderivative. (x + 1)/(x² - x + 1) Let u = x - 1/2 du = dx Then we have: (x + 1) / (x² - x + 1) = (u + 3/2) / (u² + 3/4) Now we can integrate. ∫[(x + 1)/(x² - x + 1)]dx = ∫[(u + 3/2)/(u² + 3/4)]du = (1/2)∫[2u/(u² + 3/4)]du + ∫[(3/2)/(u² + 3/4)]du = (1/2)ln(u² + 3/4) + ∫[(3/2)/(u² + 3/4)]du For the remaining integral let (√3/2)tanθ = u (√3/2)sec²θ dθ = du __________ = (1/2)ln(u² + 3/4) + (3/2)∫{1/[(3/4)tan²θ + 3/4] [(√3/2)sec²θ dθ]} = (1/2)ln(u² + 3/4) + √3∫dθ = (1/2)ln(u² + 3/4) + (√3)θ = (1/2)ln(x² - x + 1) + (√3)arctan[(2/√3)u] + C = (1/2)ln(x² - x + 1) + (√3)arctan[(2/√3)(x - 1/2)] + C = (1/2)ln(x² - x + 1) + (√3)arctan[(2x - 1)/√3] + C

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well its the Integral of Du over u u=x^2-x+1 du=2x-1 (the derivative of u) you have x+1 and you need du. so you can multiply the integral by -1/2. (you need to multiply by negative two inside the integral and compensate with -1/2) answer: -1/2ln(x^2-x+1) + C

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∫ [(x + 1)/(x² - x + 1)] dx = since the denominator doesn't factor (in that it has no real zeros), make the top the derivative of the bottom, dividing and multiplying by 2: (1/2) ∫ [2(x + 1)/(x² - x + 1)] dx = (1/2) ∫ [(2x + 2)/(x² - x + 1)] dx = to complete the derivative on the top, add and subtract - 1: (1/2) ∫ [(2x - 1 + 1 + 2)/(x² - x + 1)] dx = (1/2) ∫ {[(2x - 1) + 3]/(x² - x + 1)} dx = break it up into: (1/2) ∫ (2x - 1) dx/(x² - x + 1) + (1/2) ∫ [3 /(x² - x + 1)] dx = (1/2) ∫ d(x² - x + 1)/(x² - x + 1) + (3/2) ∫ dx /(x² - x + 1) = (1/2) ln(x² - x + 1) + (3/2) ∫ dx /(x² - x + 1) = complete the square by breaking 1 into (1/4) + (3/4): (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{[x² - x + (1/4)] + (3/4)} = (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{[x - (1/2)]² + (3/4)} = factor (3/4) out of the bottom, so as to get it into the form [f(x)]² + 1: (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{(3/4) {(4/3)[x - (1/2)]² + 1}} = include (4/3) into the square: (1/2) ln(x² - x + 1) + (3/2) ∫ dx /{(3/4) { {(2/√3)[x - (1/2)]}² + 1} } expand the base of the square and shift (3/4) to the top (as 4/3): (1/2) ln(x² - x + 1) + (3/2) ∫ (4/3)dx /{[(2/√3)x - (1/√3)]² + 1} = split (4/3) into (2/√3)(2/√3) and pull (2/√3) out, yielding (2/√3)dx, that is d[(2/√3)x - (1/√3)]: (1/2) ln(x² - x + 1) + (3/2) ∫ (2/√3)(2/√3)dx /{[(2/√3)x - (1/√3)]² + 1} = (1/2) ln(x² - x + 1) + (3/2)(2/√3) ∫ (2/√3)dx /{[(2/√3)x - (1/√3)]² + 1} = (1/2) ln(x² - x + 1) + (√3) ∫ d[(2/√3)x - (1/√3)] /{[(2/√3)x - (1/√3)]² + 1} = that is of the form ∫ d[f(x)] /{[f(x)]² + 1} = arctan[f(x)] + C so you conclude with: ∫ [(x + 1)/(x² - x + 1)] dx = (1/2) ln(x² - x + 1) + (√3)arctan[(2/√3)x - (1/√3)] + C I hope it helps..

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