How do you integrate #(6x^5 -2x^4 + 3x^3 + x^2 - x-2)/x^3#?

Question

6496 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-29T22:46:45

#=2x^3-x^2+3x+ln|x|+1/x+1/x^2+C#

Since we're dividing by a single term, #x^3,# we may simplify the integrand as follows:

#(6x^5-2x^4+3x^3+x^2-x-2)/x^3=(6x^5)/x^3-(2x^4)/x^3+(3x^3)/x^3+x^2/x^3-x/x^3-2/x^3#

#=6x^2-2x+3+1/x-1/x^2-2/x^3#

Thus, our integral becomes

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

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

Integrate, recalling that

#intx^adx=x^(a+1)/(a+1)+C, a ne -1, int1/xdx=ln|x|+C#

#=int(6x^2-2x+3+1/x-x^-2-2x^-3)dx=6/3x^3-2/2x^2+3x+ln|x|+x^-1+2/2x^-2+C#

#=2x^3-x^2+3x+ln|x|+1/x+1/x^2+C#