How do you evaluate the indefinite integral $int (-4x^4+x^2-6)dx$ ?

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Answer 1 · 2017-02-04T06:43:31

The answer is $=-4/5x^5+1/3x^3-6x+C$

We use

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

Therefore,

$int(-4x^4+x^2-6)dx$

$=-4intx^4dx+intx^2dx-6intdx$

$=-4/5x^5+1/3x^3-6x+C$

How do you evaluate the indefinite integral int (-4x^4+x^2-6)dxint (-4x^4+x^2-6)dx?