How do you integrate int (x-9)/((x+3)(x-7)(x-5)) ∫x−9(x+3)(x−7)(x−5) using partial fractions?
1 Answer
Explanation:
As the other answer was unfortunately deleted, I will answer this question.
To compute a partial fractions decomposition means to find
(x-9)/((x+3)(x-7)(x-5)) = A / (x+3) + B / (x-7) + C / (x-5)x−9(x+3)(x−7)(x−5)=Ax+3+Bx−7+Cx−5
... multiply both sides with the denominator
<=>" " x - 9 = A(x-7)(x-5) + B(x+3)(x-5) + C (x+3)(x-7)⇔ x−9=A(x−7)(x−5)+B(x+3)(x−5)+C(x+3)(x−7)
<=>" " x - 9 = A(x^2 - 12x + 35) + B(x^2 - 2x - 15) + C (x^2 - 4x - 21)⇔ x−9=A(x2−12x+35)+B(x2−2x−15)+C(x2−4x−21)
<=>" " x - 9 = A * x^2 - A * 12x + A * 35 + B * x^2 - B * 2x - B * 15 + C * x^2 - C * 4x - C * 21⇔ x−9=A⋅x2−A⋅12x+A⋅35+B⋅x2−B⋅2x−B⋅15+C⋅x2−C⋅4x−C⋅21
... identify the
<=>" " color(red)(0 * x^2) + color(blue)(x) - 9 = color(red)(A * x^2) color(blue)(- 12A * x) + 35A + color(red)(B * x^2) color(blue)(- 2B * x) - 15B + color(red)(C * x^2) color(blue)(- 4C * x) - 21C⇔ 0⋅x2+x−9=A⋅x2−12A⋅x+35A+B⋅x2−2B⋅x−15B+C⋅x2−4C⋅x−21C
This leads us to the following linear equation system:
{ (" " (I) " "0 = " "A + B " "+C " " color(red)(x^2) " terms"), (" " (II) " "1 = -12A - 2 B " "- 4C " " color(blue)(x) " terms"), (" " (III) -9 = " "35A - 15B - 21C " constant terms") :}
Now, let's solve this linear equation system.
To eliminate
B , let's compute(IV) = 2 * (I) + (II) and(V) = 15 * (I) + (III) :
(IV) " " 1 = -10 A - 2C
(V) " " -9 = " "50 A - 6C Now, we can compute
(VI) = 5 * (IV) + (V) to eliminateA :
(VI)" " -4 = -16 C Thus, we can solve
(VI) forC and findC = 1/4 .Plugging
C into(IV) gives usA = -3/20 and finally, pluggingA andC into(I) gives usB = - 1/10 .
Now, we can perform the partial fraction decomposition and solve the integral:
int (x-9)/((x+3)(x-7)(x-5)) "d"x
= int [-3/20 * 1 / (x+3) -1/10 * 1/ (x-7) + 1/4 * 1/ (x-5) ] "d"x
= -3/20 int [ 1 / (x+3)"d"x -1/10 int 1 / (x-7) "d"x + 1/4 int 1/ (x-5) "d"x
= -3/20 ln abs (x+3) -1/10 ln abs (x-7) + 1/4 ln abs (x-5) + C
