Question

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

Answer 1

`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`

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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`

Answer 2

An answer already existed