I need some help with the following indefinite integral
$$\int\frac{dx}{2x^3-6x+4}$$
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4 Answers
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Hint: Note that $1$ is a root of the equation $2x^3-6x+4=0$ and use partial fraction decomposition.
answered Sep 6, 2014 at 14:46
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Hint; Write it as $$\int \left ( \frac {1}{18(x+2)} - \frac {1}{18(x-1)} + \frac {1}{6(x-1)^2} \right ) dx$$
It's easy now, take it from here.
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Hint: $$\Large \frac{1}{2x^3-6x+4} \equiv \frac{1}{18(x+2)}-\frac{1}{18(x-1)}+\frac{1}{6(x-1)^2}$$
You can check this identity by factorising $2x^3-6x+4$ as $2(x+2)(x-1)^2.$
answered Sep 6, 2014 at 14:54
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Hint
Partial fraction decomposition should be your friend since the roots of the denominator are $1$ and $-2$ as you should have found by simple inspection.
answered Sep 6, 2014 at 14:48