Question

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

Answer 1

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`