# How to Integrate ln(x+1)/x

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The function f(x) = ln(x+1)/x looks simple enough, but if you try to find its antiderivative using standard methods of integration such as substitution or integration by parts, you will find yourself going in circles generating more and more complicated intermediate results. It turns out that ln(x+1)/x is a non-integrable function in the sense that it doesn't have a closed form antiderivative. In other words, you cannot express its antiderivative in terms of elementary functions, and in general a definite integral of ln(x+1)/x cannot be expressed in terms of elementary functions of the limits of integration.

However, you can find two convergent series representations of the antiderivative of ln(x+1)/x that will allow you to estimate definite integrals. This tutorial shows how to find these series. The method is very similar to that of integrating the related function y = ln(x)/(x+1).

## Graph of y = ln(x+1)/x

Graph of y = ln(x+1)/x with vertical asymptote at x = -1 and horizontal asymptote at y = 0. (Created with Desmos graphing calculator.)

## Approximate Integral of ln(x+1)/x, x Between -1 and 1

When -1 < x ≤ 1, the well-known convergent Taylor series expansion of the natural logarithm function ln(x+1) is

ln(x+1) = x - (x^2)/2 + (x^3)/3 - (x^4)/4 + (x^5)/5 - (x^6)/6 + ...

Dividing every term in this series by a factor of x gives us the series expansion of ln(x+1)/x:

ln(x+1)/x = 1 - x/2 + (x^2)/3 - (x^3)/4 + (x^4)/5 - (x^5)/6 + ...

Integrating each piece of the series gives us the infinite series form of the antiderivative of ln(x+1)/x:

∫ ln(x+1)/x dx =
x - (x^2)/4 + (x^3)/9 - (x^4)/16 + (x^5)/25 - (x^6)/36 + ... + C

This series can be used to approximate definite integrals of ln(1+x)/x when the limits of integration are between -1 and 1, including 1. All you need to do is truncate it after a certain number of terms. The more terms you use, the more accurate the approximation is.

## Hey Wait! Isn't ln(x+1)/x Undefined at x = 0?

An observant student might notice that ln(x+1)/x is undefined at x = 0. Plugging 0 directly into the expression yields the indeterminate form 0/0. However, the limit of the function as x approaches 0 is equal to 1, whether you approach from the negative or positive side of the axis. This allows us to use the antiderivative approximation formula for an interval that covers x = 0.

## Approximate Integral of ln(x+1)/x, x Greater Than or Equal to 1

When x ≥ 1, the Taylor series expansion of the natural logarithm function ln(x+1) give above does not converge. However, we can apply a trick to turn it into a convergent form. The key is to use the identity

ln(1 + x) = ln(x) + ln(1 + 1/x)

which is easily derived using properties of logarithms. If x is greater than or equal to 1, then its reciprocal 1/x is less than or equal to 1. This means we can construct a convergent series for ln(1 + 1/x):

ln(1 + 1/x) = 1/x - 1/(2x^2) + 1/(3x^3) - 1/(4x^4) + 1/(5x^5) - ...

Dividing every term in this series by a factor of x gives us the series expansion of ln(1 + 1/x)/x:

ln(1 + 1/x)/x = 1/x^2 - 1/(2x^3) + 1/(3x^4) - 1/(4x^5) + 1/(5x^6) - ...

Now we can integrate to find the antiderivative of the original function ln(1+x)/x

∫ ln(x+1)/x dx
= ∫ [ln(x) + ln(1 + 1/x)]/x dx
= ∫ ln(x)/x dx + ∫ ln(1 + 1/x)/x dx
= (1/2)ln(x)^2 + ∫ [1/x^2 - 1/(2x^3) + 1/(3x^4) - 1/(4x^5) + ...] dx
= (1/2)ln(x)^2 + [ -1/x + 1/(4x^2) - 1/(9x^3) + 1/(16x^4) - ... ] + C

This expression can be used to find definite integrals of ln(1+x)/x when the limits of integration are greater than or equal to 1.

## Example 1

Let's use the first series antiderivative to estimate the integral of ln(x+1)/x from x = -1/2 to x = 1/2. For this, we can use the first series antiderivative expression truncated to six terms, i.e.,

∫ ln(x+1)/x dx ≈ x - (x^2)/4 + (x^3)/9 - (x^4)/16 + (x^5)/25 - (x^6)/36

Plugging in the limits of integration and subtracting them gives us

12911/28800 - (-5587/9600) = 3709/3600 ≈ 1.03028

With a numerical integrator, the true answer is closer to 1.03065. The approximation is pretty close. If we had used more terms of the series, say 10 instead of six, the answer would have been even more accurate.

## Example 2

Now let's use the second series antiderivative to approximate the integral of ln(x+1)/x from x = 2 to x = 3. For this problem, we can truncate the series to six terms again, i.e.,

∫ ln(x+1)/x dx
≈ (1/2)ln(x)^2 - 1/x + 1/(4x^2) - 1/(9x^3) + 1/(16x^4) - 1/(25x^5) + 1/(36x^6)

Plugging in the limits of integration and subtracting gives us

(1/2)ln(3)^2 - 811007/2624400 - (1/2)ln(2)^2 + 12911/28800 ≈ 0.50252

With a numerical integrator the true answer is approximately 0.50263. Again, the approximation series gives an answer that is very close to the actual answer.

## Some Interesting Exact Definite Integrals of ln(x+1)/x

Even though over arbitrary intervals f(x) = ln(x+1)/x is non-integrable in terms of elementary functions, there are some definite integrals that can be expressed in terms of elementary functions of π and ln(2), the natural logarithm of 2.

For example, the integral of ln(x+1)/x from x = -1 to x = 0 is equal to (1/6)π^2. The integral evaluated over the interval x = -1/2 to x = 0 is (1/12)π^2 - (1/2)ln(2)^2.

The image below shows several more examples of exact definite integrals. These integrals must be worked out using more advanced math and analytic number theory.

## Related Functions Without Closed-Form Antiderivatives

The function f(x) = ln(x+1)/x is similar to other functions that cannot be integrated in terms of elementary functions, for example, sin(x)/x and cos(x)/x. In fact, many functions of the form f(x) = g(x)/x cannot be integrated exactly when g(x) is a logarithmic, exponential, or trigonometric function. Some examples include

• (e^x)/x
• tan(x)/x
• arctan(x)/x
• arcsin(x)/x
• arccos(x)/x
• arctan(x)/x

However, not all functions of the form f(x) = g(x)/x are non-integrable. The following all have closed form antiderivatives

• ln(x)/x
• sqrt(x^2 + 1)/x
• ln(ln(x))/x
• (e^x)(1 - 1/x)/x

0

2

9

4

1

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