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

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

How to Integrate 1/(x^4 + 1)

How to Integrate 1/(x^3 + 1) and 1/(x^3 - 1)

How to Integrate 1/(ax^2 + bx + c)

## 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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