How do you find the indefinite integral of #int (-4x^-4-20x^-5) dx#?

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Answer 1 · 2016-09-09T21:24:00

#=1/3x^-4(4x+15)+C#.

We use the Standard Form # : intt^ndt=t^(n+1)/(n+1), n!=-1#.

Hence, #int(-4x^-4-20x^-5)dx#

#int-4x^-4dx-int20x^-5#

#-4intx^-4dx-20intx^-5dx#

#=-4(x^(-4+1))/(-4+1)-20(x^(-5+1))/(-5+1)#

#4/3x^-3+5x^-4#

#=1/3x^-4(4x+15)+C#.