I am interested in finding an exact expression for the integral $$\int \frac{\ln(x^2+1)dx}{x}$$ I start by using the transformation $w=\ln(x^2+1)$ leading to $e^{w}dw=2xdx$. Unfortunately, I couldn't get rid of $x$ in there: $$\int \frac{we^w}{2x^2}dw$$ and I'm not interested in results with an infinite series.

I would like to know a suitable substitution that may involve only one variable leading to an expression less complicated for integration. Thanks for your help.

  • $\begingroup$ It is $$-\frac{\text{Li}_2\left(-x^2\right)}{2}$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 27 at 14:01
  • $\begingroup$ yes this is what wolfram is showing...can you explain what Li is? $\endgroup$ – Marvel Maharrnab Apr 27 at 14:03
  • $\begingroup$ Thanks. However, this kind of expression is what I want to avoid. Thus, I need alternate expression. $\endgroup$ – M.D. Apr 27 at 14:04
  • $\begingroup$ @MarvelMaharrnab This special function is called Dilogarithm, or Spence's Function. $\endgroup$ – mrtaurho Apr 27 at 14:11
  • $\begingroup$ @MarvelMaharrnab, sure this is what I got when I tried it on wolfram. And this $L_{i_n}(y)$ is defined as a polylog function $L_{i_n}(y)=\sum_{m=1}^{\infty}\frac{y^m}{m^n}$, for $n=1,2,\dots$. This is what I am avoiding because it would add extra difficulty to implement to plot numerically. $\endgroup$ – M.D. Apr 27 at 14:12

This is not possible by using elementary functions yet alone. The Dilogarithm, or Spence's Function is capable of providing an anti-derivative. Note that the aforementioned function my defined as integral

$$\operatorname{Li}_2(x):=-\int\frac{\log(1-x)}x\mathrm dx$$

Enforcing the substitution $x^2=-t$ within your given integral gives us

\begin{align*} \int\frac{\log(1+x^2)}x\mathrm dx&=\int\frac{\log(1+x^2)}x\frac{2x}{2x}\mathrm dx\\ &=\frac12\int\frac{\log(1-t)}t\mathrm dt\\ &=-\frac12\operatorname{Li}_2(t)\\ &=-\frac12\operatorname{Li}_2(-x^2) \end{align*}

$$\therefore~\int\frac{\log(1+x^2)}x\mathrm dx~=~-\frac12\operatorname{Li}_2(-x^2)$$


To remove the ugly "$-x^2$".

\begin{align}F(x)&=\int_0^x \frac{\ln(t^2+1)dt}{t}\,dt\\ &=\frac{1}{2}\int_0^x \frac{2t\ln(t^2+1)dt}{t^2}\,dt\\ \end{align} Perform the change of variable $y=t^2$, \begin{align}F(x)&=\frac{1}{2}\int_0^{x^2}\dfrac{\ln(1+t)}{t}\,dt\\ &=\frac{1}{2}\Big[\ln t\ln(1+t)\Big]_0^{x^2}-\frac{1}{2}\int_0^{x^2} \frac{\ln t}{1+t}\,dt\\ &=\ln x\ln(1+x^2)-\frac{1}{2}\int_0^{x^2} \frac{\ln t}{1+t}\,dt\\ &=\ln x\ln(1+x^2)-\frac{1}{2}\int_0^{x^2}\dfrac{\ln t}{1-t}\,dt+\int_0^{x^2}\frac{t\ln t}{1-t^2}\,dt\\ \end{align} In the last integral perform the change of variable $y=t^2$, \begin{align}F(x)&=\ln x\ln(1+x^2)-\frac{1}{2}\int_0^{x^2}\dfrac{\ln t}{1-t}\,dt+\frac{1}{4}\int_0^{x^4}\dfrac{\ln t}{1-t}\,dt\end{align}

In the first integral perform the change of variable $y=\dfrac{t}{x^2}$,

In the second integral perform the change of variable $y=\dfrac{t}{x^4}$,

\begin{align}F(x)&=\ln x\ln(1+x^2)-\frac{x^2}{2}\int_0^{1}\dfrac{\ln(tx^2)}{1-tx^2}\,dt+\frac{x^4}{4}\int_0^{1}\dfrac{\ln(tx^4)}{1-tx^4}\,dt\\ &=\ln x\ln(1+x^2)-x^2\ln x\int_0^{1}\dfrac{1}{1-tx^2}\,dt+x^4\ln x\int_0^{1}\dfrac{1}{1-tx^4}\,dt-\\ &\frac{x^2}{2}\int_0^{1}\dfrac{\ln t}{1-tx^2}\,dt+\frac{x^4}{4}\int_0^{1}\dfrac{\ln t}{1-tx^4}\,dt\\ &=\boxed{\frac{1}{2}\text{Li}_2(x^2)-\frac{1}{4}\text{Li}_2(x^4)} \end{align}


For $0\leq x<1$, $\text{Li}_2$ is defined by, \begin{align}\text{Li}_2(x)=\sum_{n=1}^{\infty} \dfrac{x^n}{n^2}\end{align}

For $0<x<1$, \begin{align}\int_0^1 \dfrac{\ln t}{1-tx}\,dt=\dfrac{\text{Li}_2(x)}{x}\end{align}

Proof: \begin{align}\int_0^1 \dfrac{\ln t}{1-tx}\,dt&=\int_0^1 \left(\ln t\sum_{n=0}^\infty (tx)^n\right)\,dt\\ &=\sum_{n=0}^{\infty} x^n\int_0^1 t^n\ln t\,dt\\ &=-\sum_{n=0}^{\infty} \frac{x^n}{(n+1)^2}\\ &=-\frac{1}{x}\sum_{n=0}^{\infty} \frac{x^{n+1}}{(n+1)^2}\\ &=-\dfrac{\text{Li}_2(x)}{x} \end{align}

Because, for $n\geq 0$, integer \begin{align}\int_0^1 t^n\ln t\,dt=-\frac{1}{(n+1)^2}\end{align}


You could try and use the taylor series centred around $a=1$ to find a series for this integral. $$I=\int\frac{\ln(x^2+1)}{x}dx=\frac12\int\frac{\ln(u)}{u-1}du$$ and we can say that: $$f(u)=\ln(u)$$ and by following the pattern of differentiating this function: $$f(u)=\ln(u)$$ $$f'(u)=0!(-1)^1x^{-1}$$ $$f''(u)=1!(-1)^2x^{-2}$$ etc. so we can say that: $$\frac{d^nf(u)}{du}=(n-1)!(-1)^{n-1}u^{-n}$$ and if $u=1$ then this $u$ term will disappear, giving: $$\ln(u)=\sum_{n=1}^\infty\frac{(u-1)^n(-1)^{n+1}}{n}$$ and so we can say that: $$I=\frac12\int\sum_{n=1}^\infty\frac{(-1)^{n+1}(u-1)^{n-1}}{n}du$$ $$I=\frac12\int\sum_{n=0}^\infty\frac{(-1)^n(u-1)^n}{n+1}du$$ $$I=\sum_{n=0}^\infty\frac{(-1)^n(u-1)^{n+1}}{(n+1)^2}+C$$ $$I=\sum_{n=0}^\infty\frac{(-1)^nx^{2(n+1)}}{(n+1)^2}+C$$


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