How do you integrate f(x)=1/((x-2)(x-5)(x+3))f(x)=1(x−2)(x−5)(x+3) using partial fractions?
1 Answer
Explanation:
Since the factors on the denominator are linear then the numerators of the partial fractions will be constants , say A , B and C
rArr 1/((x-2)(x-5)(x+3)) = A/(x-2) + B/(x-5) + C/(x+3) ⇒1(x−2)(x−5)(x+3)=Ax−2+Bx−5+Cx+3 now, multiply through by (x-2)(x-5)(x+3)
so 1 = A(x-5)(x+3) + B(x-2)(x+3) + C(x-2)(x-5) ................................(1)
We now have to find the values of A , B and C. Note that if x = 2 , the terms with B and C will be zero. If x = 5 , the terms with A and C will be zero and if x = -3, the terms with A and B will be zero.
Making use of this fact , we obtain.let x = 2 in (1) : 1 = -15A
rArr A = -1/15 ⇒A=−115 let x = 5 in (1) : 1 = 24B
rArr B = 1/24 ⇒B=124 let x = -3 in (1) : 1 = 40C
rArr C = 1/40 ⇒C=140
rArr 1/((x-2)(x-5)(x+3)) = (-1/15)/(x-2) + (1/24)/(x-5) + (1/40)/(x+3)⇒1(x−2)(x−5)(x+3)=−115x−2+124x−5+140x+3 Integral now becomes.
-1/15intdx/(x-2) + 1/24intdx/(x-5) + 1/40intdx/(x+3) −115∫dxx−2+124∫dxx−5+140∫dxx+3
= 1/24ln|x-5| + 1/40ln|x+3| - 1/15ln|x-2| + c =124ln|x−5|+140ln|x+3|−115ln|x−2|+c
