Integral $\int_0^1\frac{x^9\left(x^4+x^2-x-1-5\ln x\right)}{\left(x^{10}-1\right)\ln x}\mathrm dx$ A friend of mine sent me an integral that she had not been able to crack, and me neither. It comes with a result, but without a proof (I suppose it originated in some math contest). Could you please suggest an approach to prove the result?
$$\int_0^1\frac{x^9\left(x^4+x^2-x-1-5\ln x\right)}{\left(x^{10}-1\right)\ln x}\mathrm dx=\frac12\gamma+\frac{11}5\ln2-\frac54\ln5+\frac12\ln\pi-\frac12\ln\phi,$$
where $\gamma$ is the Euler–Mascheroni constant, and $\phi$ is the golden ratio.
 A: Let's denote the integral in question as
$$I=\int_0^1\frac{x^9\left(x^4+x^2-x-1-5\ln x\right)}{\left(x^{10}-1\right)\ln x}dx.\tag1$$
Changing the variable $x=y^{1/10}$ and renaming $y$ back to $x$ we get
$$I=\int_0^1\frac{x^{2/5}+x^{1/5}-x^{1/10}-1-\ln\sqrt x}{(x-1)\ln x}dx.\tag2$$
Some elementary transformations show that
$$I=\mathcal{J}(2/5)+\mathcal{J}(1/5)-\mathcal{J}(1/10),\tag3$$
where we introduced notation
$$\mathcal{J}(q)=\int_0^1\frac{x^q-1-q\,\ln x}{(x-1)\ln x}dx.\tag4$$
The integral $\mathcal{J}(q)$ can be evaluated as follows:
$$\begin{align}\mathcal{J}(q)=\int_0^1\int_0^q\frac{x^p-1}{x-1}dp\,dx=\int_0^q\underbrace{\int_0^1\frac{x^p-1}{x-1}dx}_{\text{DLMF 5.9.16}}\,dp\\=\int_0^q H_p\,dp=q\cdot\gamma+\ln\Gamma(q+1),\end{align}\tag5$$
where $H_p$ are harmonic numbers: $H_p$$\,=\,$$\gamma$$\,+\,$$\psi_0$$(p+1)$, and $\psi_0$ is the digamma function: $\psi_0(x)=\frac{d}{dx}\ln\,$$\Gamma$$(x)$. Let me mention that the formula DLMF 5.9.16 becomes particularly obvious for positive integer $p$, when $H_p=\sum_{n=1}^pn^{-1}$.
Pluging $(5)$ back into $(3)$, we get
$$I=\frac12\gamma+\ln\frac45+\ln\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac25\right)}{\Gamma\left(\frac1{10}\right)}.\tag6$$
From the formula $(74)$ on this MathWorld page we know that
$$\frac{\Gamma\left(\frac15\right)\Gamma\left(\frac25\right)}{\Gamma\left(\frac1{10}\right)}=\frac{\sqrt[5]2\,\sqrt\pi}{\sqrt[4]5\,\sqrt\phi}.\tag7$$
(see the paper Raimundas Vidūnas, Expressions for values of the gamma function for a proof).
Making use of this formula, we get the final result
$$I=\frac12\gamma+\frac{11}5\ln2-\frac54\ln5+\frac12\ln\pi-\frac12\ln\phi.\tag8$$
