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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