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How do you find the indefinite integral of #int (-4/x^3-8/x^5) dx#?

1 Answer

Nov 27, 2016

The answer is #=2/x^2+2/x^4+C#

Explanation:

We use #intx^ndx=x^(n+1)/(n+1)+C (n!=-1)#

So, #int(-4/x^3-8/x^5)dx=int(-4x^(-3)-8x^(-5))dx#

#=-4*x^(-3+1)/(-3+1)-8*x^(-5+1)/(-5+1)+C#

#=2/x^2+2/x^4+C#

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