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

The function f(x) = ln(x)/(x+1) looks like it should be easy to integrate, but when you try to find its antiderivative using standard tools such as substitution or integration by parts, you will never be able to reduce it to a known form. It turns out that ln(x)/(x+1) 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)/(x+1) cannot be written in terms of elementary functions of the limits of integration.

However, you can find two convergent series representations of the antiderivative of ln(x)/(x+1) that will let you to estimate definite integrals. This tutorial shows how to find these series. You might notice that the method of finding an approximate integral of ln(x)/(x+1) is very similar to that of the closely related function **y = ln(x+1)/x**.

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

## Approximate Integral of ln(x)/(x+1), x Between 0 and 1

To work out a convergent series antiderivative for ln(x)/(x+1), we can start with the convergent series representation of 1/(x+1) for x between 0 and 1. This is

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

Multiplying every term by ln(x) gives us a convergent series for ln(x)/(x+1) for x between 0 and 1:

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

To construct the series integral of ln(x)/(x+1), we simply integrate each term in the series above. To do this, we can use the antiderivative formula for functions of the form (x^a)ln(x), which is worked out in the linked tutorial.

∫ ln(x)/(x+1) dx

= ∫ ln(x) dx - ∫ ln(x)x dx + ∫ ln(x)x^2 dx - ∫ ln(x)x^3 dx + ∫ ln(x)x^4 dx - ...

= ln(x)[x - (x^2)/2 + (x^3)/3 - (x^4)/4 + ...]

- [x - (x^2)/4 + (x^3)/9 - (x^4)/16 + ...] + C

The series x - (x^2)/2 + (x^3)/3 - ... is the convergent series expansion of ln(x+1) for x ≤ 1, which means we can simplify the integral expression to

∫ ln(x)/(x+1) dx

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

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

If you truncate this series antiderivative and use a finite number of terms, you will get a good approximation of the integral of ln(x)/(x+1) for values of x between 0 and 1, including 1. An example is worked out after the next section.

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

Using a similar process, we can construct a series antiderivative of ln(x)/(x+1) when x is greater than 1. First, we need to find a convergent series for 1/(x+1) when x is greater than one. This is

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

Multiplying each term by ln(x) gives us a convergent series for ln(x)/(x+1):

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

Taking the integral of each term gives us the antiderivative series

∫ ln(x)/(x+1) dx

= ∫ ln(x)/x dx - ∫ ln(x)/x^2 dx + ∫ ln(x)/x^3 dx - ∫ ln(x)/x^4 dx + ...

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

+ [1/x - 1/(4x^2) + 1/(9x^3) - 1/(16x^4) + ...] + C

As in the previous section, the form of this series can be simplified if you recognize that the series 1/x - 1/(2x^2) + 1/(3x^3) - ... is a convergent series for ln(1 + 1/x) when x ≥ 1. Using this fact we can further simplify the antiderivative series.

∫ ln(x)/(x+1) dx

= (1/2)ln(x)^2 + ln(x)*ln(1 + 1/x)

+ [1/x - 1/(4x^2) + 1/(9x^3) - 1/(16x^4) ...] + C

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

+ [1/x - 1/(4x^2) + 1/(9x^3) - 1/(16x^4) + ...] + C

= ln(x)*ln(x+1) - (1/2)ln(x)^2 + [1/x - 1/(4x^2) + 1/(9x^3) - ...] + C

= ln(x)*ln(x+1) - (1/2)ln(x)^2 - [-1/x + 1/(4x^2) - 1/(9x^3) + ...] + C

This infinite series antiderivative converges for all values of x greater than 1, and for x = 1 as well. To estimate a definite integral of ln(x)/(x+1) over an interval [p, q] with 1 ≤ p < q, simply truncate the series and use a finite number of terms.

## Example 1: Integral of ln(x)/(x+1) Over [1/2, 1]

Let's find the approximate integral of ln(x)/(x+1) between x = 1/2 and x = 1. To do this, we use the first series antiderivative of ln(x)/(x+1) truncated to 10 series terms. In other words, we use the equation

∫ ln(x)/(x+1) dx ≈ ln(x)ln(x+1) - x + (x^2)/4 - (x^3)/9 + (x^4)/16 - (x^5)/25 + (x^6)/36 - (x^7)/49 + (x^8)/64 - (x^9)/81 + (x^10)/100

If you plug in the limits of integration and subtract, you get -0.0885. With a numerical integrator, the true value of the integral is about -0.0930058. This estimate with the 10-term truncation is pretty crude, but if we had truncated it at 20 terms the series integral formula would have given a better estimate.

## Example 2

Now let's find the approximate integral of ln(x)/(x+1) between x = 2 and x = 3. To do this, we use the second series antiderivative of ln(x)/(x+1) truncated to 10 series terms. In other words, we use the equation

∫ ln(x)/(x+1) dx ≈ ln(x)ln(x+1) - (1/2)ln(x)^2 + 1/x - 1/(4x^2) + 1/(9x^3) - 1/(16x^4) + 1/(25x^5) - 1/(36x^6) + 1/(49x^7) - 1/(64x^8) + 1/(81x^9) - 1/(100x^10)

If you plug in the limits of integration and subtract, you get 0.2589. With a numerical integrator, the true value of the integral is also about -0.2589. This estimate with the 10-term truncation is pretty good.

## An Exact Definite Integral of ln(x)/(x+1)

The integral of ln(x)/(x+1) from x = 0 to x = 1 converges to an exact value of -(π^2)/12. The definite integral is negative because the curve lies below the x-axis over the interval (0, 1).

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