How do you integrate #y=(-2-3x)/(x+x^2+2x^3)# using the quotient rule?

Question

1217 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-30T00:42:46

#(dy)/(dx)=(-2-4x-15x^2-12x^3)/(x+x^2+2x^3)^2#

#y=(-2-3x)/(x+x^2+2x^3)#

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

#(dy)/(dx)=(-2-4x-12x^2-3x-6x^2-18x^3+3x+3x^2+6x^3)/(x+x^2+2x^3)^2#

#(dy)/(dx)=(-2-4x-15x^2-12x^3)/(x+x^2+2x^3)^2#

REMEMBER, the quotient rule is

#y=u/v#

#(dy)/(dx)=(v'u-vu')/v^2#