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How do you find the general indefinite integral of #(13x^2+12x^-2)dx#?

1 Answer

Feb 15, 2015

The answer is: #13/3x^3-12x^-1+c#

Using this integral:

#int[f(x)]^nf'(x)dx=[f(x)]^(n+1)/(n+1)+c#.

So:

#int(13x^2+12x^-2)dx=13x^(2+1)/(2+1)+12x^(-2+1)/(-2+1)+c=#

#=13/3x^3-12x^-1+c#.

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