How do you integrate #int (-9/x^4)dx#?

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Answer 1 · 2017-02-03T07:06:13

The answer is #=3/x^3+C#

We need

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

Here,

#int-9/x^4dx=-9intx^-4dx#

#=-9x^(-4+1)/(-4+1)+C#

#=-9/-3*1/x^3*C#

#=3/x^3+C#