# How to Integrate 1/(x^4 - 1)

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Finding the integral of 1/(x^4 - 1) is much easier than finding the integral of 1/(x^4 + 1) because the quartic polynomial x^4 - 1 is readily factorable into lower degree polynomials with integer coefficients. This makes it easier to use the method of partial fractions to transform the rational function 1/(x^4 - 1) into the sum of simpler rational functions. The antiderivative of 1/(x^4 - 1) allows you to find the area under the curve of the graph and solve some types of separable differential equations. If you want help integrating similar rational functions, see also

## Step 1: Factoring x^4 - 1

The quartic polynomial x^4 - 1 factors into one quadratic and two linear polynomials.

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

This means we can use the method of partial fractions to rewrite 1/(x^4 - 1) as

(Ax + B)/(x^2 + 1) + C/(x + 1) + D/(x - 1)

Once we know the values of A, B, C, and D, it will be easy to integrate because the antiderivatives of these simple rational functions are logarithms and inverse tangents.

## Step 2: Finding A, B, C, and D

Since we want to find the values of A, B, C, and D such that

1/(x^4 - 1) = (Ax + B)/(x^2 + 1) + C/(x + 1) + D/(x - 1)

we can start by making the right-hand side of the equation a single rational function, i.e., finding the common denominator. Doing so gives us a denominator of x^4 - 1 and a numerator of

Ax^3 - Ax + Bx^2 - B + Cx^3 - Cx^2 + Cx - C + Dx^3 + Dx^2 + Dx + D

This numerator must equal 1, the numerator of 1/(x^4 - 1). This means the coefficients of x^3, x^2, and x must equal 0. Grouping together like terms gives us a set of four linear equations in four unknowns:

A + C + D = 0
B - C + D = 0
-A + C + D = 0
-B - C + D = 1

The solution to this linear system is A = 0, B = -1/2, C = -1/4, and D = 1/4.

## Step 3: Integrating

Now that we have shown that the quartic rational function 1/(x^4 - 1) is equivalent to the rational function sum

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

we can integrate this expression term by term. The first is an inverse tangent function and the last two are natural logs.

∫ (-1/2)/(x^2 + 1) dx = (-1/2)arctan(x) + C

∫ (-1/4)/(x + 1) dx = (-1/4)Ln |x + 1| + C

∫ (1/4)/(x - 1) dx = (1/4)Ln |x - 1| + C

Putting everything together, we get the antiderivative of 1/(x^4 - 1).

∫ 1/(x^4 - 1) dx =
(-1/2)arctan(x) - (1/4)Ln |x + 1| + (1/4)Ln |x - 1| + C
= (-1/2)arctan(x) + (1/4)Ln |(x-1)/(x+1)| + C

## Applications to Other Integrals in Calculus

In another tutorial we found the antiderivative of 1/(x^4 + 1), which was a more complicated process due to the more difficult factorization of x^4 + 1. Knowing both the integral of 1/(x^4 - 1) and 1/(x^4 + 1) we can also find the integral of 1/(x^8 - 1). This is because x^8 - 1 = (x^4 + 1)(x^4 - 1), and

1/(x^8 - 1) = (1/2)/(x^4 - 1) - (1/2)/(x^4 + 1)

Therefore, we have

∫ 1/(x^8 - 1) dx = [ ∫ 1/(x^4 - 1) dx - ∫ 1/(x^4 + 1) dx ] / 2

## Area Under the Curve y = 1/(x^4 - 1)

The area under the curve 1/(x^4 - 1) from x = sqrt(3) to x = ∞ is found by plugging these endpoints into the antiderivative expression and subtracting. When dealing with an infinite point, you actually take the limit. This gives you

lim(x→∞) [(-1/2)arctan(x) + (1/4)Ln |(x-1)/(x+1)|]
= (-1/2)(π/2) + (1/4)Ln(1)
= -π/4

Plugging x = sqrt(3) into the antiderivative gives you

(-1/2)arctan(sqrt(3)) + (1/4)Ln(2 - sqrt(3))
= -π/6 + (1/4)Ln(2 - sqrt(3))

Subtracting the two gives the total area

-π/4 + π/6 - (1/4)Ln(2 - sqrt(3))
= -π/12 - (1/4)Ln(2 - sqrt(3))
= (1/4)Ln(2 + sqrt(3)) - π/12
≈ 0.06744

## Another Example

Solve the separable differential equation dy/dx = 4(y - 1)/(x^4 - 1). Using the technique of separation of variables we have

(1/4)/(y - 1) dy = 1/(x^4 - 1) dx

Integrating both sides gives us

(1/4)Ln |y-1| = (-1/2)arctan(x) + (1/4)Ln |(x-1)/(x+1)| + C

Ln |y-1| = -2arctan(x) + Ln |(x-1)/(x+1)| + C

By making each side the exponent of the natural logarithm base e we can solve this equation to obtain y as a function of x.

y - 1 = e^[-2arctan(x) + Ln |(x-1)/(x+1)| + C]

y - 1 = C*[(x-1)/(x+1)]e^(-2arctan(x))

y = C*[(x-1)/(x+1)]e^(-2arctan(x)) + 1

Compare this to the similar differential equation dy/dx = 4(y-1)/x^4, whose solution is the function y(x) = 1 + C*e^(-4/(3x^3)). Just a small change in the denominator makes a large difference in the functional form of the solution. If, for the sake of example, we add the condition lim(x→∞) y(x) = 7.4463 to our original differential equation, then we can find the value of C.

lim(x→∞) C*[(x-1)/(x+1)]e^(-2arctan(x)) + 1 = 7.4463

lim(x→∞) C*[(x-1)/(x+1)]e^(-2arctan(x)) = 6.4463

C*1*e^(-pi) = 6.4463

C = 149.1718

A graph of this function is shown below.

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