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Integral of Ln(1 + x^3) and Ln(x^3 - 1)

Updated on October 10, 2015
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TR Smith is a product designer and former teacher who uses math in her work every day.

The graph above shows the functions y = Ln(x^3 + 1), y = Ln(x^3) = 3*ln(x), and y = Ln(x^3 - 1). As you can see in the figure above, the graphs of the functions Ln(x^3 + 1) and Ln(x^3 - 1) are almost indistinguishable from the graph of Ln(x^3) = 3*Ln(x) for large values of x. However, they look quite different for values of x close to 0 and 1. While the antiderivative of 3*Ln(x) is the simple function 3x*Ln(x) - 3x, the antiderivatives of Ln(x^3 + 1) and Ln(x^3 - 1) are much more complicated.

To integrate Ln(x^3 + 1), it is necessary to factor the polynomial x^3 + 1. Then, using a combination of integration by parts and algebra, you can transform the integral into simpler quadratic rational functions. The process to integrate Ln(x^3 - 1) is nearly identical. This calculus tutorial will work out the integral of Ln(x^3 + 1) step by step so you can see how to integrate the logarithm of any factorable cubic function. See also, How to Integrate 1/(x^3 + 1) and 1/(x^3 - 1).


Step 1: Ln(x^3 + 1) = Ln(x + 1) + Ln(x^2 - x + 1)

The first step to integrate Ln(x^3 + 1) is to use properties of logarithms and polynomial factoring to simplify the integral. Since x^3 + 1 = (x + 1)(x^2 - x + 1) and Log(A*B) = Log(A) + Log(B), we have

∫ Ln(x^3 + 1) dx = ∫ Ln(x + 1) dx + ∫ Ln(x^2 - x + 1) dx

The integral of Ln(x + 1) is easy, its antiderivative is (x+1)*Ln(x+1) - (x+1) + C, which you can figure out using the tutorial for integrating Ln(x). To find the antiderivative of Ln(x^2 - x + 1), we will use integration by parts.


Step 2: Integral of Ln(x^2 - x + 1)

There are two ways to find the antiderivative of Ln(x^2 - x + 1). One is to rewrite the function as Ln[(x - 1/2)^2 + 3/4] and make the substitution w = x - 1/2, then use the tutorial for finding the antiderivative of Ln(x^2 + 1). Or we can use integration by parts on the integral

∫ Ln(x^2 - x + 1) dx

using the assignment u = Ln(x^2 - x + 1) and dv = dx, which gives us du = (2x - 1)/(x^2 - x + 1) dx and v = x. This gives us

∫ Ln(x^2 - x + 1) dx =

x*Ln(x^2 - x + 1) - ∫ (2x^2 - x)/(x^2 - x + 1) dx

Now we just need to find the integral of the rational function (2x^2 - x)/(x^2 - x + 1).


Step 3: Integral of (2x^2 - x)/(x^2 - x + 1)

We can use algebra to rewrite this rational function in a more convenient form.

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

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

Each of these can easily be integrated. The first is linear, the second is logarithmic, and the third is inverse trigonometric.

∫ 2 dx = 2x + C

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

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

The last integral can be found applying the technique discussed in How to Integrate 1/(ax^2 + bx + c). Now all that's left to do is put all the integrals from the three steps together.


Step 4: Putting It All Together

Combining the integrals found in Steps 1 - 3 above, we get the complete integral of Ln(x^3 + 1).

∫ Ln(1 + x^3) dx =

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

By combining like terms, using properties of logarithms, and rolling the constant terms together, we can put this expression in a more compact form:

∫ Ln(1 + x^3) dx =

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

How to Integrate Ln(x^3 - 1)

Once you have seen how to integrate Ln(x^3 + 1), you can easily find the antiderivative of Ln(x^3 - 1) since the steps are basically the same. First we factor the polynomial x^3 - 1 to rewrite the function as L(x - 1) + Ln(x^2 + x + 1). Next, we integrate both logarithms. The second logarithm can be resolved with integration by parts, which results in another rational function. The integral his rational function also works out to be a linear, logarithmic, and inverse trigonometric function. Putting everything together gives you

∫ Ln(x^3 - 1) dx =

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

Using the same simplification techniques, we can also put this function in a more compact form.

∫ Ln(x^3 - 1) dx =

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

Logarithms of Polynomials

The above technique can be used to integrate any polynomial logarithm so long as you can factor the polynomial into linear and quadratic terms. The work of integrating Ln(x^3 + 1) and Ln(x^3 - 1) also allows you to integrate Ln(x^6 - 1), since x^6 - 1 = (x^3 + 1)(x^3 - 1). And using the facts that

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

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

you can also find the antiderivatives of Ln(x^4 + 1) and Ln(x^4 - 1).

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