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I am trying to find this math by integrating by parts but I am unable to do it. $$\int \frac{\ln x}{1-x^2}dx$$

Andrei
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Sujit23
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1 Answers1

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First use partial fraction decomposition to split.

$$\int \frac{\ln(x)}{1-x^2}dx = \frac{1}{2}\int \frac{\ln(x)}{1+x}dx + \frac{1}{2}\int \frac{\ln(x)}{1-x}dx$$

Then using integration by parts for each. For the first, choose to integrate $\frac{1}{1+x}dx$ and to differentiate $\ln(x)$. Similarly for the other, integrate $\frac{1}{1-x}dx$ and differentiate $\ln(x)$.

$$\frac{1}{2}\ln(x)\ln(1+x) - \frac{1}{2}\int \frac{\ln(1+x)}{x}dx -\frac{1}{2}\ln(x)\ln(1-x) + \frac{1}{2}\int \frac{\ln(1-x)}{x}dx$$

Now, for the two integrals left, both can be manipulated to be in the form of $-\int \frac{\ln(1-x)}{x}dx = \mathrm{Li_2}(x) + C$, where $\mathrm{Li_2}(x)$ is the dilogarithm function. Thus we have,

$$\int \frac{\ln(x)}{1-x^2}dx = \frac{1}{2}\ln(x)\ln(1+x) -\frac{1}{2}\ln(x)\ln(1-x) + \frac{1}{2}\mathrm{Li_2}(-x) - \frac{1}{2}\mathrm{Li_2}(x) + C$$.

Hopefully answering your question.

MichaelCatliMath
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  • Can you explain me more about dilogarithm function. Thank in advance. – Sujit23 Apr 11 '20 at 05:16
  • Sorry for the late response. The dilogarithm function is a special function that exists for many reasons, but one of them being because the integral $\int \frac{\ln(1-x)}{x}dx$ is nonelementary. Note that it’s pretty close to $\int \frac{\ln(x)}{x}dx$, which is elementary and can be integrated via the substitution $u=\ln(x)$. Note that elementary basically means/implies that its integral can be expressed in standard mathematical notation, and nonelementary being the opposite. There’s more here about the specific dilogarithm function, and its family, the polylogarithm: – MichaelCatliMath Apr 30 '20 at 02:49
  • https://en.m.wikipedia.org/wiki/Polylogarithm – MichaelCatliMath Apr 30 '20 at 02:50