How do you integrate ∫3x2−6x+8 using partial fractions?
1 Answer
Mar 11, 2016
Explanation:
Begin by factorising the denominator.
x2−6x+8=(x−2)(x−4) Since these factors are linear then the numerators of the partial fractions will be constants , say A and B.
⇒3(x−2)(x−4)=Ax−2+Bx−4 Multiply through by (x - 2 )(x - 4 )
3 = A(x - 4 ) + B(x - 2 ) .......................................(1)
The aim now is to find the value of A and B. Note that if x = 4 , the term with A will be zero and if x = 2 the term with B will be zero.
let x = 4 in (1) : 3 = 2B⇒B=32
let x = 2 in (1) : 3 = -2A⇒A=−32
⇒3(x−2)(x−4)=32x−4−32x−2 and the integral becomes :
32∫dxx−4−32∫dxx−2
=32ln|x−4|−32|x−2|+c
