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

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Rational functions with a cubic in the denominator can be hard to integrate if you can't factor the cubic by hand into three linear factors, or one linear factor and an irreducible quadratic factor. But the functions 1/(1 + x^3) and 1/(x^3 - 1) can easily be factored and fall into the latter category. That is

x^3 + 1 = (x + 1)(x^2 - x + 1) and
x^3 - 1 = (x - 1)(x^2 + x + 1)

This allows you to use the integration technique of partial fractions to decompose the cubic rational into the sum of a simple linear rational function and a quadratic rational function. The resulting antiderivative contains both the natural log and the inverse tangent function.

## Graphs of y = 1/(x^3 +1) and y = 1/(x^3 - 1)

Click thumbnail to view full-size
Graph of y = 1/(x^3 + 1)
Graph of y = 1/(x^3 - 1)
Graphs of y = 1/(x^3 + 1) [blue], y = 1/(x^3 - 1) [green], y = 1/x^3 [black]
Graphs of y = 1/(x^3 + 1) [orange] and y = 1/(x^3 + 8) [violet]

## Integral of 1/(x^3 + 1) -- Part 1

The first thing we do is apply partial fractions, that is, we find constants A, B, and C such that

1/(x^3 + 1) = A/(x+1) + (Bx + C)/(x^2 - x + 1).

thereby breaking up the original cubic denominator into a linear factor and an irreducible quadratic factor. To solve for A, B, and C, we multiply both sides of the above equation by x^3 + 1 to clear the denominator of each term. This gives us

1 = A(x^2 - x + 1) + (Bx + C)(x + 1)
1 = (A + B)x^2 + (-A + B + C)x + (C + A)

The left side of the equation can be thought of as the quadratic polynomial 0x^2 + 0x + 1, so we need to find the values of A, B, and C such that

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

Using a 3-by-3 matrix equation to solve this system yields A = 1/3, B = -1/3, and C = 2/3. This gives us the partial fractions decomposition

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

## Integral of 1/(x^3 + 1) -- Part 2

The next step is to integrate the partial fractions expression. The term (1/3)/(x + 1) is easy to integrate and its antiderivative is (1/3)*Ln|x + 1| + k. Now we need to work out the antiderivative of the quadratic rational function (-x/3 + 2/3)/(x^2 - x + 1). To do this, we rewrite it as

(-x/3 + 1/6)/(x^2 - x + 1) + (1/2)/(x^2 - x + 1)

Let's integrate the first half of this new expression using u-substitution with u = x^2 - x + 1 and du = (2x - 1) dx. This implies that (-x/3 + 1/6) dx = (-1/6) du. Thus we have

∫ (-x/3 + 1/6)/(x^2 - x + 1) dx
= ∫ (-1/6)/u du
= (-1/6)Ln(u) + k
= (-1/6)Ln(x^2 - x + 1) + k

And now we integrate the second half of the expression using completion of the square.

∫ (1/2)/(x^2 - x + 1) dx
= (1/2)*∫ 1/[(x - 1/2)^2 + 3/4] dx

Using the substitution u = x - 1/2 and du = dx we get

(1/2)*∫ 1/(u^2 + 3/4) du
= (1/2) * [2 / sqrt(3)] * arctan[ 2u / sqrt(3) ] + k
= [1/sqrt(3)] * arctan[(2x - 1)/sqrt(3)] + k

Putting everything together, we finally get the complete expression for the antiderivative of 1/(x^3 + 1):

∫ 1/(x^3 + 1) dx
= (1/3)Ln|x + 1| - (1/6)Ln(x^2 - x + 1) + sqrt(1/3)*arctan[(2x-1)/sqrt(3)] + k

## Integral of 1/(x^3 - 1)

Now that we've completely integrated 1/(x^3 + 1), finding the antiderivative of 1/(x^3 - 1) is much easier since the steps are almost identical. As in the previous sections, we first find constants A, B, and C such that

1/(x^3 - 1) = A/(x - 1) + (Bx + C)/(x^2 + x + 1)

Sparing the details, the coefficients work out to be A = 1/3, B = -1/3, and C = -2/3. This gives us

1/(x^3 - 1) = (1/3)/(x - 1) + (-x/3 - 2/3)/(x^2 + x + 1)
= (1/3)/(x - 1) + (-x/3 - 1/6)/(x^2 + x + 1) + (-1/2)/(x^2 + x + 1)

Now we can integrate each term in the sum.

• ∫ (1/3)/(x-1) dx
= (1/3)Ln|x - 1| + k

• ∫ (-x/3 - 1/6)/(x^2 + x + 1) dx
= ∫ (-1/6)/u du
= (-1/6)Ln(u) + k
= (-1/6)Ln(x^2 + x + 1) + k

• ∫ (-1/2)/(x^2 + x + 1) dx
= (-1/2)*∫ 1/[(x + 1/2)^2 + 3/4] dx
= (-1/2) * [2/sqrt(3)] * arctan[(2x + 1)/sqrt(3)] + k
= -sqrt(1/3)*arctan[(2x+1)/sqrt(3)] + k

Putting everything together gives you

∫ 1/(x^3 - 1) dx
= (1/3)Ln|x - 1| - (1/6)Ln(x^2 + x + 1) - sqrt(1/3)*arctan[(2x+1)/sqrt(3)] + k

## Example Integration Problem

Find the area under the curve y = 1/(x^3 + 1) from x = 0 to x = infinity. This region is shaded in yellow in the graph below.

Since the antiderivative of f(x) = 1/(x^3 + 1) is known, we simply need to plug in the endpoints x = 0 and x = infinity and subtract. To plug in x = infinity we need to use limits.

lim(x→) (1/6)Ln[(x+1)(x+1)/(x^2 - x + 1)] + sqrt(1/3)*arctan[(2x-1)/sqrt(3)]

= (1/6)*Ln(1) + sqrt(1/3)*arctan(∞)

= (1/6)*0 + sqrt(1/3)*(π/2)

= π/[2*sqrt(3)]

The endpoint x = 0 can be plugged in directly to give us

(1/6)Ln(1/1) + sqrt(1/3)*arctan[-1/sqrt(3))]

= (1/6)*0 - sqrt(1/3)*(π/6)

= -π/[6(sqrt(3)]

Finally, if we subtract the two values we get the area under the curve from 0 to infinity:

π/[2(sqrt(3)] - (-π)/[6(sqrt(3)] = 2π/[3(sqrt(3)]

## Integrals of 1/(x^3 + a^3)

Using the factorization x^3 + a^3 = (x + a)(x^2 - ax + a^2), you can also work out the integral of 1/(x^3 + a^3) when 'a' is a positive or negative real number. The integral works out to be

∫ 1/(x^3 + a^3) dx =
(1/(6a^2))Ln[(x^2 + 2ax + a^2)/(x^2 - ax + a^2)]
+ (1/(sqrt(3)*a^2))arctan[(2x - a)/(sqrt(3)*a)] + C

## Related Integrals

Here are four more calculus integration tutorials for closely related functions.

## Integral of 1/(x^6 - 1)

Given the relation

1/(x^6 - 1) = (1/2)[ 1/(x^3 - 1) - 1/(x^3 + 1) ]

you can use the antiderivatives of 1/(x^3 + 1) and 1/(x^3 - 1) to find the antiderivative of 1/(x^6 - 1). This works out to

∫ 1/(x^6 - 1) dx
= (1/12)Ln[(x^4 - 3x^3 + 4x^2 - 3x + 1)/(x^4 + 3x^3 + 4x^2 + 3x + 1)]
- (sqrt(3)/6)arctan[(2x+1)/sqrt(3)] - (sqrt(3)/6)arctan[(2x-1)/sqrt(3)] + C

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