Integral $\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$

Question

A while ago I encountered this integral $$I=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx$$ To be fair I spent some time with it and solved it in a heuristic way, I want to avoid that way so I won't show that approach, but the result I got is $\frac{\pi}{6} \ln(2+\sqrt 3)$ so I bet it can be shown in a nice way.
Solving other integrals I also encountered this one: $$J=\int_0^\infty \frac{\ln(1+x^2+x^4)}{1+x^2}dx=\pi \ln(2+\sqrt 3)$$ Which is pretty easy to compute, so most of my time I tried to show that $J=6I$, however without explictly evaluating them I had no luck.
Also I tried to use partial fractions: $$I=\frac12 \left(\int_0^1 \frac{\ln(1+x)}{x^2+\sqrt 3x +1}dx - \int_0^1 \frac{\ln(1+x)}{x^2-\sqrt 3 x+1}dx\right) $$ Considering: $$K(t) =\int_0^1 \frac{\ln(1+x)}{x^2-2\cos(t)x+1}dx$$ We have that $I=\frac12 \left(K\left(\frac{5\pi}{6}\right)-K\left(\frac{\pi}{6}\right)\right) $ and since: $$\frac{\sin t}{x^2-2x\cos t+1}=\frac{1}{2i}\left(\frac{e^{it}}{1-xe^{it}}-\frac{e^{-it}}{1-xe^{-it}}\right)=\Im \left(\frac{e^{it}}{1-xe^{it}}\right)=$$ $$=\sum_{n=0}^{\infty} \Im\left(x^n e^{i(n+1)t}\right)=\sum_{n=0}^\infty x^n\sin((n+1)t)$$ $$\small \Rightarrow K(t)=\frac12 \left(\frac{1}{\sin \left(\frac{5\pi}{6}\right)}\sum_{n=0}^\infty \sin\left(\frac{5\pi}{6} (n+1)t \right) + \frac{1}{\sin \left(\frac{\pi}{6}\right)}\sum_{n=0}^\infty \sin\left(\frac{\pi}{6} (n+1)t \right)\right)\int_0^1 x^n \ln(1+x)dx$$ $$=2\sum_{n=0}^\infty \sin\left(\frac{\pi}{6} (n+1)t \right)\int_0^1 x^n \ln(1+x)dx$$ Now I don't know how to deal with the integral and the sum combined.

Answer 1

\begin{align}J&=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx\\ &=\int_0^1\int_0^1 \frac{x(x^2+1)}{(x^4-x^2+1)(1+xt)}\,dt\,dx\\ &=-\int_0^1\int_0^1 \frac{t(t^2+1)}{(t^4-t^2+1)(1+xt)}\,dt\,dx+\int_0^1\int_0^1 \frac{t^3+t^2x-\sqrt{3}t^2-\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2-\sqrt{3}x+1)}\,dt\,dx+\\ &\int_0^1\int_0^1 \frac{t^3+t^2x+\sqrt{3}t^2+\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2+\sqrt{3}x+1)}\,dt\,dx\\ &=-J+\int_0^1\int_0^1 \frac{t^3+t^2x-\sqrt{3}t^2-\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2-\sqrt{3}x+1)}\,dt\,dx+\\ &\int_0^1\int_0^1 \frac{t^3+t^2x+\sqrt{3}t^2+\sqrt{3}tx+t+x}{2(t^4-t^2+1)(x^2+\sqrt{3}x+1)}\,dt\,dx\\ \end{align}

Since,

\begin{align}A_1&=\int_0^1 \frac{t}{t^4-t^2+1}\,dt\\ &=\frac{1}{\sqrt{3}}\left[\arctan\left(\frac{2t^2-1}{\sqrt{3}}\right)\right]_0^1\\ &=\frac{\pi}{3\sqrt{3}}\\ A_3&=\int_0^1 \frac{t^3}{t^4-t^2+1}\,dt\\ &=\frac{1}{12}\left[2\sqrt{3}\arctan\left(\frac{2t^2-1}{\sqrt{3}}\right)+3\ln(t^4-t^2+1)\right]_0^1\\ &=\frac{\pi}{6\sqrt{3}}\\ B_1&=\int_0^1 \frac{1}{x^2-\sqrt{3}x+1}\,dx\\ &=2\Big[\arctan\left(2x-\sqrt{3}\right)\Big]_0^1\\ &=\frac{5\pi}{6}\\ B_2&=\int_0^1 \frac{1}{x^2+\sqrt{3}x+1}\,dx\\ &=2\Big[\arctan\left(2x+\sqrt{3}\right)\Big]_0^1\\ &=\frac{\pi}{6}\\ C_1&=\int_0^1 \frac{x}{x^2-\sqrt{3}x+1}\,dx\\ &=\Big[\frac{1}{2}\ln\left(x^2-\sqrt{3}x+1\right)+\sqrt{3}\arctan\left(2x-\sqrt{3}\right)\Big]_0^1\\ &=\frac{1}{2}\ln\left(2-\sqrt{3}\right)+\frac{5}{4\sqrt{3}}\pi\\ C_2&=\int_0^1 \frac{x}{x^2+\sqrt{3}x+1}\,dx\\ &=\Big[\frac{1}{2}\ln\left(x^2+\sqrt{3}x+1\right)-\sqrt{3}\arctan\left(2x+\sqrt{3}\right)\Big]_0^1\\ &=\frac{1}{2}\ln\left(2+\sqrt{3}\right)-\frac{1}{4\sqrt{3}}\pi\\ A_2&=\int_0^1 \frac{t^2}{t^4-t^2+1}\,dt\\ &=\int_0^1 \frac{t^2+1}{t^4-t^2+1}\,dt-\int_0^1 \frac{1}{t^4-t^2+1}\,dt\\ &=\left[\arctan\left(\frac{x}{1-x^2}\right)\right]_0^1-\frac{1}{2\sqrt{3}}\int_0^1 \frac{x}{x^2+\sqrt{3}x+1}\,dx+\frac{1}{2\sqrt{3}}\int_0^1 \frac{x}{x^2-\sqrt{3}x+1}\,dx-\\ &\frac{1}{2}\int_0^1 \frac{1}{x^2+\sqrt{3}x+1}\,dx-\frac{1}{2}\int_0^1 \frac{1}{x^2-\sqrt{3}x+1}\,dx\\ &=\frac{\pi}{2}-\frac{1}{2\sqrt{3}}C_2+\frac{1}{2\sqrt{3}}C_1-\frac{1}{2}B_2-\frac{1}{2}B_1\\ &=\frac{\pi}{4}-\frac{1}{2\sqrt{3}}\ln\left(2+\sqrt{3}\right)\\ A_0&=\int_0^1 \frac{1}{t^4-t^2+1}\,dt\\ &=\int_0^1 \frac{1+t^2}{t^4-t^2+1}\,dt-\int_0^1 \frac{t^2}{t^4-t^2+1}\,dt\\ &=\left[\arctan\left(\frac{x}{1-x^2}\right)\right]_0^1-A_2\\ &=\frac{\pi}{4}+\frac{1}{2\sqrt{3}}\ln\left(2+\sqrt{3}\right)\\ \end{align}

then,

\begin{align} 2J&=\left(\frac{1}{2}A_3B_1+\frac{1}{2}A_2C_1-\frac{\sqrt{3}}{2}A_2B_1-\frac{\sqrt{3}}{2}A_1C_1+\frac{1}{2}A_1B_1+\frac{1}{2}A_0C_1\right)+\\ &\left(\frac{1}{2}A_3B_2+\frac{1}{2}A_2C_2+\frac{\sqrt{3}}{2}A_2B_2+\frac{\sqrt{3}}{2}A_1C_2+\frac{1}{2}A_1B_2+\frac{1}{2}A_0C_2\right)\\ \end{align}

Since,

\begin{align}\frac{1}{2}A_3B_1&=\frac{5\pi^2}{72\sqrt{3}}\\ \frac{1}{2}A_2C_1&=\frac{5\pi^2}{32\sqrt{3}}+\frac{\pi}{16}\ln\left(2-\sqrt{3}\right)-\frac{1}{8\sqrt{3}}\ln\left(2-\sqrt{3}\right)\ln\left(2+\sqrt{3}\right)-\frac{5\pi}{48}\ln\left(2+\sqrt{3}\right)\\ -\frac{\sqrt{3}}{2}A_2B_1&=\frac{5\pi}{24}\ln\left(2+\sqrt{3}\right)-\frac{5\pi^2}{16\sqrt{3}}\\ -\frac{\sqrt{3}}{2}A_1C_1&=-\frac{\pi}{12}\ln\left(2-\sqrt{3}\right)-\frac{5\pi^2}{24\sqrt{3}}\\ \frac{1}{2}A_1B_1&=\frac{5\pi^2}{36\sqrt{3}}\\ \frac{1}{2}A_0C_1&=\frac{\pi}{16}\ln\left(2-\sqrt{3}\right)+\frac{5\pi^2}{32\sqrt{3}}+\frac{1}{8\sqrt{3}}\ln\left(2-\sqrt{3}\right)\ln\left(2+\sqrt{3}\right)+\frac{5\pi}{48}\ln\left(2+\sqrt{3}\right)\\ \frac{1}{2}A_3B_2&=\frac{\pi^2}{72\sqrt{3}}\\ \frac{1}{2}A_2C_2&=\frac{\pi}{12}\ln\left(2+\sqrt{3}\right)-\frac{\pi^2}{32\sqrt{3}}-\frac{1}{8\sqrt{3}}\ln^2\left(2+\sqrt{3}\right)\\ \frac{\sqrt{3}}{2}A_2B_2&=\frac{\pi^2}{16\sqrt{3}}-\frac{\pi}{24}\ln\left(2+\sqrt{3}\right)\\ \frac{\sqrt{3}}{2}A_1C_2&=\frac{\pi}{12}\ln\left(2+\sqrt{3}\right)-\frac{\pi^2}{24\sqrt{3}}\\ \frac{1}{2}A_1B_2&=\frac{\pi^2}{36\sqrt{3}}\\ \frac{1}{2}A_0C_2&=\frac{1}{8\sqrt{3}}\ln^2\left(2+\sqrt{3}\right)+\frac{\pi}{24}\ln\left(2+\sqrt{3}\right)-\frac{\pi^2}{32\sqrt{3}}\\ \end{align}

Therefore,

\begin{align}2J&=\frac{\pi}{24}\ln\left(2-\sqrt{3}\right)+\frac{3\pi}{8}\ln\left(2+\sqrt{3}\right)\end{align}

Since,

\begin{align}\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\end{align}

Therefore,

\begin{align}2J&=-\frac{\pi}{24}\ln\left(2+\sqrt{3}\right)+\frac{3\pi}{8}\ln\left(2+\sqrt{3}\right)\\ &=\frac{\pi}{3}\ln\left(2+\sqrt{3}\right) \end{align}

Thus,

\begin{align}\boxed{J=\frac{\pi}{6}\ln\left(2+\sqrt{3}\right)} \end{align}

Answer 2

Let $J(a) =\int_0^1 \frac{1}{1+x} \cot^{-1}\frac {2x\sin a}{1-x^2}\ dx$, along with $J(0)=\frac\pi2\ln2$ \begin{align} J’(a) &= \int_0^1 \frac{2\cos a \ (x^2-x)}{(x^2+1)^2-(2x\cos a)^2}dx =\frac12\left( a \ln\tan\frac a2\right)’ - \frac\pi4\tan\frac a2 \end{align} Then \begin{align} &\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx\\ \overset{ibp}= &\int_0^1 \frac{1}{1+x} \cot^{-1}\frac x{1-x^2}dx =J\left(\frac\pi6\right)= J(0)+\int_0^{\frac\pi6}J’(a)da \\ = &\ \frac\pi2 \ln2-\frac\pi4 \int_0^{\frac\pi6}\tan\frac a2 da+\frac12\int_0^{\frac\pi6} d\left( a\ln\tan\frac a2\right)\\ =&\ \frac\pi2\ln2+\frac\pi2\ln\cos\frac\pi{12}+\frac\pi{12}\ln\tan\frac\pi{12} = \frac\pi6\ln(2+\sqrt3) \end{align}

Answer 3

$$I=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx\overset{\large{x=\frac{1-t}{1+t}}}=4\int_0^1 \frac{(t^2+1)\left(\ln 2 - \ln(1+t)\right)}{t^4+14t^2+1}dt$$ $$\overset{IBP}=\frac{\pi}{2}\ln 2- (2+\sqrt 3)\left( \int_0^1 \frac{\ln(1+t)}{t^2+(2+\sqrt 3)^2}dt +\int_0^1 \frac{\ln(1+t)}{(2+\sqrt 3)^2 t^2+1}dt\right)$$ $$\int_0^1 \frac{\ln(1+t)}{t^2+(2+\sqrt 3)^2}dt=\int_0^\infty \frac{\ln(1+t)}{t^2+(2+\sqrt 3)^2}dt-\int_0^1 \frac{\ln(1+t)-\ln t}{(2+\sqrt 3)^2t^2+1}dt$$ $$\Rightarrow I=\frac{\pi}{2}\ln 2 -(2+\sqrt 3)\left(\int_0^\infty \frac{\ln(1+t)}{t^2+(2+\sqrt 3)^2}dt+\int_0^1 \frac{\ln t}{(2+\sqrt 3)^2t^2+1}dt\right) $$ By substituting $t=(2+\sqrt 3) x$ respectively $(2+\sqrt 3)t =x$ we get: $$I=\frac{\pi}{2}\ln 2 -\left(\int_0^\infty \frac{\ln\left(1+(2+\sqrt 3)x\right)}{1+x^2}dx+\int_0^{2+\sqrt 3} \frac{\ln\left(\frac{x}{2+\sqrt 3}\right)}{1+x^2}dx\right)$$ $$=\frac{\pi}{2}\ln 2 + \ln(2+\sqrt 3) \arctan( x)\bigg|_0^{2+\sqrt 3}-\int_0^\infty \frac{\ln\left(1+(2+\sqrt 3)x\right)}{1+x^2}dx-\int_0^{2+\sqrt 3} \frac{\ln\left(x\right)}{1+x^2}dx$$ $$=\frac{\pi}{2}\ln 2 + \frac{5\pi}{12}\ln(2+\sqrt 3)-(J_1(2+\sqrt 3)+J_2(2+\sqrt 3))$$ $$J_1(a)=\int_0^\infty \frac{\ln(1+ax)}{1+x^2}dx\Rightarrow J_1'(a)=\int_0^\infty \frac{x}{(1+x^2)(1+ax)}dx=$$ $$=\frac{a}{1+a^2} \int_0^\infty \frac{1}{1+x^2}dx+\frac{1}{1+a^2}\int_0^\infty \left(\frac{x}{1+x^2}-\frac{a}{1+ax}\right)dx=$$ $$=\frac{\pi}{2}\frac{a}{1+a^2} +\frac{1}{1+a^2}\ln\left(\frac{\sqrt{1+x^2}}{1+ax}\right)\bigg|_0^\infty=\frac{\pi}{2}\frac{a}{1+a^2}-\frac{\ln a}{1+a^2}$$ $$J_1(0)=0 \Rightarrow J_1(2+\sqrt 3) =\int_0^{2+\sqrt 3}\left(\frac{\pi}{2}\frac{a}{1+a^2}-\frac{\ln a}{1+a^2}\right)da$$ $$J_1(2+\sqrt 3)+J_2(2+\sqrt 3) =\int_0^{2+\sqrt 3}\frac{a}{1+a^2}da-\int_0^{2+\sqrt 3}\frac{\ln a}{1+a^2}da+\int_0^{2+\sqrt 3}\frac{\ln a}{1+a^2}da$$ $$\Rightarrow I=\frac{\pi}{2}\ln 2 + \frac{5\pi}{12}\ln(2+\sqrt 3)+\frac{\pi}{2}\int_0^{2+\sqrt 3}\frac{a}{1+a^2}da=$$ $$= \frac{\pi}{2}\ln 2 + \frac{5\pi}{12}\ln(2+\sqrt 3) +\frac{\pi}{4}\ln(4(2+\sqrt 3))$$ $$\Rightarrow I=\int_0^1 \frac{(x^2+1)\ln(1+x)}{x^4-x^2+1}dx= \frac{\pi}{6}\ln(2+ \sqrt 3)$$

Answer 4

I thought about the comment you left the other day, and it has led me to this alternative approach, which is more direct than my other answer. Thanks a lot, I have learned so much from you!

Let $\omega=\sqrt[3]{-1}$, and let Ti2 denote the inverse tangent integral function:

\begin{align*} I&=\int_0^1\frac{\left(1+x^2\right)\log(1+x)}{x^4-x^2+1}dx\\ &=\int_0^1\left(\frac{\omega}{1+\omega^2 x^2}+\frac{\bar{\omega}}{1+\bar{\omega}^2 x^2}\right)\log\left(1+x\right)dx\\ &=\left.\left(\arctan(\omega x)+\arctan(\bar{\omega}x)\right)\log(1+x)\right|_0^1-\int_0^1\frac{\arctan(\omega x)+\arctan(\bar{\omega} x)}{1+x}dx\\ &=\frac{\pi}{2}\log 2-\operatorname{Ti_2}(\omega,\omega)-\operatorname{Ti_2}(\bar{\omega},\bar{\omega}) \end{align*}

A result (3.27) from Polylogarithms and Associated Functions reads: $$\operatorname{Ti_2}(a,a)+\operatorname{Ti_2}(a^{-1},a^{-1})=\frac{\pi}{2}\log\left(\frac{2}{\sqrt{1+a^2}}\right)+\arctan(a)\log(a) $$

Thus letting $a=\omega$, we get:

$$I=\frac{\pi}{2}\log\left(\sqrt{1+\omega^2}\right)-\arctan(\omega)\log(\omega)=\frac{\pi}{6}\log\left(2+\sqrt{3}\right) $$

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This approach also produces a general result, for when $-\frac{\pi}{2}<\varphi<\frac{\pi}{2}$:

$$\int_0^1\frac{\left(1+x^2\right)\log(1+x)}{x^4+2\cos(2\varphi)x^2+1}dx=\frac{\pi}{8\cos\varphi}\log(2\cos\varphi)+\frac{\varphi}{4\cos\varphi}\log\tan\left(\frac{\varphi}{2}+\frac{\pi}{4}\right) $$

Answer 5

Here is a hint:

For some constants $p$,$q$ consider the integral

$$L_{p,q} = \int_0^1 \frac{\ln(1+px)}{1+qx} dx$$

and do an integration by parts (Focus on Integration of $\frac{1}{1+qx}$, while the other factor will only be differentiated). Thus:

$$L_{p,q} = [\ln(1+px)\ln(1+qx)/q]_0^1 - \frac{p}{q} \int_0^1 \frac{\ln(1+qx)}{1+px} dx $$$$= [\ln(1+px)\ln(1+qx)/q]_0^1 - \frac{p}{q} L_{q,p} = \ln(1+p)\ln(1+q)/q - \frac{p}{q} L_{q,p}$$

This is only a System of linear algebraic equations for the $L_{q,p}$. Identify the parameters and solve the system of two algebraic equations (one for the integral of desire and the other one with $p$, $q$ reversed).