Final answer to the problem
$\frac{\left(x^3-1\right)\left(x+1\right)}{\left(x^3-2x^2-3x\right)\left(x^2+x-2\right)\left(x^2+x+1\right)\left(6x+x^2-x^3\right)}$
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Step-by-step Solution
How should I solve this problem?
Find the derivative using the product rule Find the derivative using the definition Find the derivative using the quotient rule Find the derivative using logarithmic differentiation Find the derivative Integrate by partial fractions Product of Binomials with Common Term FOIL Method Integrate by substitution Integrate by parts
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Intermediate steps
$\frac{\left(x^3-1\right)\left(x+1\right)}{\left(x^3-2x^2-3x\right)\left(x^2+x-2\right)\left(x^2+x+1\right)\left(6x+x^2-x^3\right)}$
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Final answer to the problem
$\frac{\left(x^3-1\right)\left(x+1\right)}{\left(x^3-2x^2-3x\right)\left(x^2+x-2\right)\left(x^2+x+1\right)\left(6x+x^2-x^3\right)}$