$$\int\frac{1-7x}{x^2-5x+6}\,dx$$ $$\int\frac{x-1}{x^3+x^2}\,dx$$ I've been trying for like an hour to no avail. Can somebody solve them step-by-step so I can correct what I'm doing wrong?
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Hint: partial fractions. And did you try them in that one hour? – Parcly Taxel Apr 16 '20 at 10:32
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I think I've tried that, probably still made a mistake – David Apr 16 '20 at 10:41
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Factorise the denominator, and then express the numerator as a linear combination of those factors – Dhanvi Sreenivasan Apr 16 '20 at 10:57
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Have you see this site (https://www.integral-calculator.com/)? For things like this its pretty helpfull – user6767509 Apr 16 '20 at 10:58
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Thanks to you both! I'll try – David Apr 16 '20 at 10:59
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The partial fraction decompositions of the two integrands are as follows: $$\frac{1-7x}{x^2-5x+6}=\frac{13}{x-2}-\frac{20}{x-3}$$ $$\frac{x-1}{x^3+x^2}=\frac2x-\frac1{x^2}-\frac2{x+1}$$ Now you can just apply standard integration formulas to each component and sum them up.
Parcly Taxel
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