How do you integrate #int (4x^3+6x^2-1)dx#?

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Answer 1 · 2016-12-08T08:34:56

We can use the power rule for integration all the way across that integrand.

#int(4x^3+6x^2-1)dx#

Keep in mind that:

#intx^ndx=x^(n+1)/(n+1)+c#

So, unsimplified:

#=4*(x^(3+1))/(3+1)+6*(x^(2+1))/(2+1)-1(x)+c#

#=x^4+2x^3-x+c#