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

1 Answer
Mar 25, 2018

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

Explanation:

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.