How do you integrate int (1-2x^2)/((x+1)(x-6)(x-7)) ∫1−2x2(x+1)(x−6)(x−7) using partial fractions?
2 Answers
= -1/56 ln abs(x+1)+71/7 ln abs(x-6)-97/8 ln abs(x-7)+C=−156ln|x+1|+717ln|x−6|−978ln|x−7|+C
Explanation:
int (1-2x^2)/((x+1)(x-6)(x-7)) dx∫1−2x2(x+1)(x−6)(x−7)dx
= int (-1/56(1/(x+1))+71/7(1/(x-6))-97/8(1/(x-7))) dx=∫(−156(1x+1)+717(1x−6)−978(1x−7))dx
= -1/56 ln abs(x+1)+71/7 ln abs(x-6)-97/8 ln abs(x-7)+C=−156ln|x+1|+717ln|x−6|−978ln|x−7|+C
Where did those coefficients come from?
(1-2x^2)/((x+1)(x-6)(x-7)) = a/(x+1)+b/(x-6)+c/(x-7)1−2x2(x+1)(x−6)(x−7)=ax+1+bx−6+cx−7
We can calculate
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
An answer already existed
