How do you integrate int (x^3 - 2) / (x^4 - 1) using partial fractions?

1 Answer
Jul 12, 2017

1/4ln|((x^2+1)(x+1)^3)/(x-1)|+tan^-1(x) + C

Explanation:

First, let's factor x^4-1.

x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)

Using partial fraction separation, we can say that:

(x^3-2)/((x-1)(x+1)(x^2+1)) = A/(x-1) + B/(x+1)+(Cx+D)/(x^2+1)

x^3-2 = A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x+1)(x-1)

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Both the A and Cx+D terms have a (x+1) factor, so we can let x equal -1 and solve for B.

(-1)^3-2 = A(0)(2)+B(-2)(2)+(-C+D)(0)(-2)

-3 = -4B

3/4 = B

Now we can substitute this into the original equation.

x^3-2 = A(x+1)(x^2+1)+3/4(x-1)(x^2+1)+(Cx+D)(x+1)(x-1)

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Substitute x=1 to get rid of (Cx+D).

1-2 = A(2)(2) + 3/4(0)(2) + (C+D)(2)(0)

-1 = 4A

-1/4 = A

Now substitute this into the original equation.

x^3-2 = -1/4(x+1)(x^2+1)+3/4(x-1)(x^2+1)+(Cx+D)(x+1)(x-1)

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Plug in x=0 to get rid of C.

-2 = -1/4(1)(1)+3/4(-1)(1)+(0C+D)(1)(-1)

-2 = -1/4 - 3/4 - D

-1 = -D

1 = D

Now substitute this into the original equation:

x^3-2 = -1/4(x+1)(x^2+1)+3/4(x-1)(x^2+1)+(Cx+1)(x+1)(x-1)

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Now we can pretty much plug in any number to solve for C. Let's do x=2 since that won't make any terms zero.

2^3-2 = -1/4(3)(5)+3/4(1)(5)+(2C+1)(3)(1)

6 = -15/4+15/4 + (6C+3)

6 = 6C+3

3=6C

1/2 = C

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So we have:

(x^3-2)/((x-1)(x+1)(x^2+1)) = (-1/4)/(x-1) + (3/4)/(x+1)+(1/2x+1)/(x^2+1)

All that is left to do is take the integral of this function.

int[(-1/4)/(x-1) + (3/4)/(x+1)+(1/2x+1)/(x^2+1)]dx

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

We can integrate the first two pretty easily, and then we can split the numerator of the last integral into two.

= -1/4ln|x-1|+3/4ln|x+1|+int(1/2x)/(x^2+1)dx + int1/(x^2+1)dx

Use the substitution u = (x^2+1), " "du = 2xdx:

= -1/4ln|x-1|+3/4ln|x+1|+int(1/4)/udu + int1/(x^2+1)dx

= -1/4ln|x-1|+3/4ln|x+1|+1/4ln(u) + tan^-1(x)

And finally simplify everything (don't forget +C !)

= -1/4ln|x-1|+1/4ln|(x+1)^3| + 1/4ln|x^2+1| + tan^-1(x)

= 1/4ln|((x^2+1)(x+1)^3)/(x-1)|+tan^-1(x) + C

Final Answer