How do you integrate int (x^3 - 2) / (x^4 - 1) using partial fractions?
1 Answer
Explanation:
First, let's factor
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
(-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)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute
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)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Plug in
-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
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
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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
= -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
= -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