How do you calculate the antiderivative of $(-2x^3+14x^9)/(x^-2)$ ?

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Answer 1 · 2018-03-25T19:12:10

$-1/3x^6+7/6x^12+C$

Recall that $(a+b)/c=a/c+b/c$

Thus, we can rewrite $(-2x^3+14x^9)/x^-2$ as $(-2x^3)/x^-2+(14x^9)/x^-2=-2x^(3-(-2))+14x^(9-(-2))=-2x^5+14x^11$

So, we want

$int(-2x^5+14x^11)dx$

Split this up, as we can split up sums or differences when integrating:

$int-2x^5dx+int14x^11dx$

Factor out the constants:

$-2intx^5dx+14intx^11dx$

Now, recall $intx^adx=x^(a+1)/(a+1)+C$ where $C$ is the constant of integration, just an arbitrary constant.

So, integrating, we get

$(-2x^6)/6+(14x^12)/12=-1/3x^6+7/6x^12+C$

Yes, we would technically have two constants of integration as we had two integrals, but we absorb them all into one constant.

How do you calculate the antiderivative of (-2x^3+14x^9)/(x^-2)(-2x^3+14x^9)/(x^-2)?