Question

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

Answer 1

`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