How do you integrate int (x)/((x^2-4)(x-3) dx using partial fractions?

1 Answer
Steve M
Feb 1, 2017

int x/((x^2-4)(x-3)) \ dx = -1/2ln |x-2| - 1/10ln|x+2| + 3/5ln|x-3| + C

Explanation:

We want to evaluate the integral:

int x/((x^2-4)(x-3)) \ dx

We can factorise the integrand:

x/((x^2-4)(x-3)) = x/((x-2)(x+2)(x-3))

And so the integrand will have a partial fraction decomposition of the form:

x/((x-2)(x+2)(x-3)) -= A/(x-2) + B/(x+2) + C/(x-3)

Leading to:

x -= A(x+2)(x-3) + B(x-2)(x-3) + C(x-2)(x+2)

Subs x=2 \ \ \ \ \=> 2 \ \ \ \ \ = A(4)(-1) \ \ \ \ \ => A=-1/2
Subs x=-2 => -2 = B(-4)(-5) => B=-1/10
Subs x=3 \ \ \ \ \ => 3 \ \ \ \ \ = C(1)(5) \ \ \ \ \ \ \ \ \ => C=3/5

So the partial fraction decomposition of the integrand is:

x/((x-2)(x+2)(x-3)) -= (-1/2)/(x-2) + (-1/10)/(x+2) + (3/5)/(x-3)

So the integral becomes:

int x/((x^2-4)(x-3)) \ dx = int \ (-1/2)/(x-2) - (1/10)/(x+2) + (3/5)/(x-3) \ dx
" " = -1/2ln |x-2| - 1/10ln|x+2| + 3/5ln|x-3| + C

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