How do you integrate #int (1-2x^2)/((x+1)(x-6)(x-7)) # using partial fractions?

2 Answers
Oct 14, 2016

#int (1-2x^2)/((x+1)(x-6)(x-7)) dx#

#= -1/56 ln abs(x+1)+71/7 ln abs(x-6)-97/8 ln abs(x-7)+C#

Explanation:

#int (1-2x^2)/((x+1)(x-6)(x-7)) dx#

#= int (-1/56(1/(x+1))+71/7(1/(x-6))-97/8(1/(x-7))) dx#

#= -1/56 ln abs(x+1)+71/7 ln abs(x-6)-97/8 ln abs(x-7)+C#

#color(white)()#
Where did those coefficients come from?

#(1-2x^2)/((x+1)(x-6)(x-7)) = a/(x+1)+b/(x-6)+c/(x-7)#

We can calculate #a, b, c# using Heaviside's cover up method:

#a = (1-2(color(blue)(-1))^2)/(color(red)(cancel(color(black)(((color(blue)(-1))+1))))((color(blue)(-1))-6)((color(blue)(-1))-7)) = (-1)/((-7)(-8)) = -1/56#

#b = (1-2(color(blue)(6))^2)/(((color(blue)(6))+1)color(red)(cancel(color(black)(((color(blue)(6))-6))))((color(blue)(6))-7)) = (-71)/((7)(-1)) = 71/7#

#c = (1-2(color(blue)(7))^2)/(((color(blue)(7))+1)((color(blue)(7))-6)color(red)(cancel(color(black)(((color(blue)(7))-7))))) = (-97)/((8)(1)) = -97/8#

Oct 14, 2016

An answer already existed