How do you integrate $\frac{x + 2}{2 x^{3} - 8 x}$ $(x+2)/(2x^3-8x)$ using partial fractions?

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Answer 1 · 2017-10-30T13:45:09

$int \ (x+2)/(2x^3-8x) \ dx = 1/4ln|(x-2)/x| + C$

We seek:

$I = int \ (x+2)/(2x^3-8x) \ dx$

We can write as:

$I = int \ (x+2)/(2x(x^2-4)) \ dx$
$\ \ = 1/2 \ int \ (x+2)/(x(x+2)(x-2) \ dx$ (removable discontinuity)
$\ \ = 1/2 \ int \ 1/(x(x-2) \ dx$

We can now decompose the integrand in to partial fractions:

$1/(x(x-2)) -= A/x + B/(x-2)$
$" " = (A(x-2)+Bx)/(x(x-2))$

Leading to the identity:

$1 -= A(x-2)+Bx$

Where $A,B$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put $x = 0 => 1=-2A => A = -1/2$
Put $x = 2 => 1=2B => B = 1/2$

So we can now write:

$I = 1/2 \ int \ (-1/2)/x + (1/2)/(x-2) \ dx$
$\ \ = 1/4 \ int \ 1/(x-2) - 1/x \ dx$

Which now consists of standard integral so we integrate to get:

$I = 1/4{ln|x-2| - ln|x| } + C$
$\ \ = 1/4ln|(x-2)/x| + C$

How do you integrate (x+2)/(2x^3-8x)x+22x3−8x(x+2)/(2x^3-8x) using partial fractions?