How do you integrate int (x^3 + 2) / (x^4 -16) using partial fractions?

1 Answer
Jun 6, 2016

3/16 ln abs(x+2) + 5/16 ln abs(x-2) + 1/4 ln(x^2+4) - 1/8 arctan (x/2) + K

Explanation:

Firstly, you need the dx at the end of the integral for formal presentation i.e. int (x^3 + 2)/(x^4-16) dx.

Secondly, you need to decompose (x^3 + 2)/(x^4-16) into partial fractions. This is made simpler by noting that

x^4-16 = (x^2-4)(x^2+4) = (x+2)(x-2)(x^2+4)

That is,

(x^3 + 2)/(x^4-16) = A/(x+2) + B/(x-2) + (Cx+D)/(x^2+4)

for real constants A, B, C and D.

Using the cover-up rule and comparison of coefficients by algebraic means, you should be able to find solve for the unknowns, that is, A = 3/16, B = 5/16, C=1/2 and D=–1/4.

Thus, (x^3 + 2)/(x^4-16) = (3/16)/(x+2) + (5/16)/(x-2) + (1/2x-1/4)/(x^2+4), and

int (x^3 + 2)/(x^4-16) dx=int ((3/16)/(x+2) + (5/16)/(x-2) + (1/2x-1/4)/(x^2+4)) dx.

The first two integrals are simple, i.e.

int (3/16)/(x+2) dx = 3/16 int 1/(x+2) dx = 3/16 ln abs(x+2)+K_1

and likewise,

int (5/16)/(x+2) dx = 5/16 ln abs(x-2)+K_2

The third part is a little tricky. By appropriate decompositions,

int (1/2x-1/4)/(x^2+4) dx = 1/4 int (2x)/(x^2+4) dx - 1/4 int 1/(x^2+4) dx = 1/4 ln(x^2+4) - 1/4(1/2)arctan (x/2) + K_3 = 1/4 ln(x^2+4) - 1/8 arctan (x/2) + K_3

Thus, putting everything together,

int (x^3 + 2)/(x^4-16) dx=3/16 ln abs(x+2) + 5/16 ln abs(x-2) + 1/4 ln(x^2+4) - 1/8 arctan (x/2) + K

where K = K_1 + K_2 + K_3.