How do you use partial fraction decomposition to decompose the fraction to integrate $\frac{3}{x^{2} - 6 x + 8}$ $3/(x^2-6x+8)$ ?

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Answer 1 · 2015-06-16T18:01:45

$3 \int \frac{1}{x^{2} - 6 x + 8} d x = 3 \int - \frac{1}{2 (x - 2)} + \frac{1}{2 (x - 4)}$ $3int1/(x^2-6x+8)dx = 3int -1/(2(x-2))+1/(2(x-4)$

$3 \int \frac{1}{x^{2} - 6 x + 8} d x$ $3int1/(x^2-6x+8)dx$

You need to factorize the denominator before using partial fraction decomposition

note : $x^{2} - 6 x + 8 = (x - 2) (x - 4)$ $x^2-6x+8 = (x - 2)(x-4)$

$3 \int \frac{1}{(x - 2) (x - 4)} d x = \frac{A}{x - 2} + \frac{B}{x - 4}$ $3int1/((x-2)(x-4))dx = A/(x-2) + B/(x-4)$

$= \frac{A (x - 4)}{(x - 2) (x - 4)} + \frac{B (x - 2)}{(x - 4) (x - 2)}$ $= (A(x-4))/((x-2)(x-4)) + (B(x-2))/((x-4)(x-2))$

$= A x - 4 A + B x - 2 B = x (A + B) - 4 A - 2 B$ $= Ax - 4A + Bx-2B = x(A+B) - 4A - 2B$

$A + B = 0$ $A + B = 0$
$- 4 A - 2 B = 1$ $-4A - 2B = 1$

So

$A = - \frac{1}{2}$ $A = -1/2$
$B = \frac{1}{2}$ $B = 1/2$

Then

$3 \int \frac{1}{x^{2} - 6 x + 8} d x = 3 \int - \frac{1}{2 (x - 2)} + \frac{1}{2 (x - 4)}$ $3int1/(x^2-6x+8)dx = 3int -1/(2(x-2))+1/(2(x-4)$

How do you use partial fraction decomposition to decompose the fraction to integrate 3x2−6x+83/(x^2-6x+8)?