Final answer to the problem
$3abc\left(31a^2b-21ab^2c^{3}-42ac^2+62b^{3}c\right)$
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Step-by-step Solution
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- Integrate by partial fractions
- Product of Binomials with Common Term
- FOIL Method
- Integrate by substitution
- Integrate by parts
- Integrate using tabular integration
- Integrate by trigonometric substitution
- Weierstrass Substitution
- Prove from LHS (left-hand side)
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1
Factor the polynomial $93a^3b^2c-63a^2b^3c^4-126a^2bc^3+186ab^4c^2$ by it's greatest common factor (GCF): $3abc$
$3abc\left(31a^2b-21ab^2c^{3}-42ac^2+62b^{3}c\right)$
Final answer to the problem
$3abc\left(31a^2b-21ab^2c^{3}-42ac^2+62b^{3}c\right)$
