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integrate 1/(x^2-x-6)
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i believe they need you to decompose the fraction, if possible
(x^2-x-6)=(x-3)(x+2) 1/(x-3)(x+2)=A/x-3+B/(x+2) A(x+2)+B(x-3)/(x^2-x-6) A(x+2)+B(x-3)=1 when x=3 A(5)=1 A=1/5 when x=-2 B(-5)=1 B=-1/5 \[1/5\int\limits_{}^{}1/(x+2)dx-1/5\int\limits_{}^{}1/(x-3)dx\] 1/5(ln(x+2))-1/5ln(x-3)+C
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