What is #int (x^3-2x^2+6x-3 ) / (-x^2+ 9 x +2 )#?

Question

What is #int (x^3-2x^2+6x-3    ) / (-x^2+  9  x  +2     )#?

Ezekiel Alexander · Accepted Answer

#int(x^3-2x^2+6x-3)/(-x^2+9x+2)dx=#
#-x^2/2-7x-71/2*ln(x^2-9x-2)#
#-(661sqrt(89))/178*ln((2x-9-sqrt89)/(2x-9+sqrt89))+C# From the given integrand, divide the numerator by the denominator. So that #(x^3−2x^2+6x−3)/(−x^2+9x+2)=-x-7-(71x+11)/(x^2-9x-2)# We have to simplify the third term, so that #(71x+11)/(x^2-9x-2)=((71/2)(2x-9+9+(2*11)/71))/(x^2-9x-2)# that is to force into the numerator a term #(2x-9)# which is the derivative of #(x^2-9x-2)#, enabling us to use #int((du)/u)=ln u+C#. We can also use #int (du)/(u^2-a^2)=1/(2a)*ln((u-a)/(u+a))+C# afterwards the continuation is #(71x+11)/(x^2-9x-2)=((71/2)(2x-9+661/71))/(x^2-9x-2)# #(71x+11)/(x^2-9x-2)=((71/2)(2x-9)+(71/2)(661/71))/(x^2-9x-2)# Now, we are ready to integrate #int(x^3−2x^2+6x−3)/(−x^2+9x+2)dx=int(-x-7-(71x+11)/(x^2-9x-2))dx# #int(x^3−2x^2+6x−3)/(−x^2+9x+2)dx=# #int(-x-7-((71/2)(2x-9)+(71/2)(661/71))/(x^2-9x-2))dx# #int(-x-7-((71/2)(2x-9))/(x^2-9x-2)-((71/2)(661/71))/(x^2-9x-2))dx# also by "completing the square" #int(-x-7-((71/2)(2x-9))/(x^2-9x-2)-((661/2))/((x-9/2)^2-(sqrt89/2)^2))dx# and #-x^2/2-7x# #-71/2*ln(x^2-9x-2)-(661sqrt(89))/178*ln((2x-9-sqrt89)/(2x-9+sqrt89))+C# God bless....I hope the explanation is useful.

Emilia Aguilar · Answer

undefined The expression $ \frac{{x^3 - 2x^2 + 6x - 3}}{{-x^2 + 9x + 2}} $ can be simplified by polynomial division. The quotient is $ x + 7 $ with a remainder of $ \frac{{-57x + 1}}{{x^2 - 9x - 2}} $.