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Find the closed form for

$$\int_{0}^{1}\left(x^k+{\ln{x}\over 1-x}\right)^2dx=f(k)\tag0$$

Setting $k=0$ then $f(0)=1$.

$f(k)$ seems to be rational numbers for all values of $k\ge0.$

We are going to prove that

$$I=\int_{0}^{1}\left(1+{\ln{x}\over 1-x}\right)^2dx=\color{red}{1}?\tag1$$

Expand $(1)$

$$I=\int_{0}^{1}\left(1+{2\ln{x}\over 1-x}+{\ln^2{x}\over (1-x)^2}\right)dx=1$$

Standard integral $$\int_{0}^{1}{\ln{x}\over 1-x}dx=-{\pi^2\over 6}$$

Then

$$I=1-{\pi^2\over 3}+\int_{0}^{1}{\ln^2{x}\over (1-x)^2}dx=1\tag2$$

$$J=\int_{0}^{1}{\ln^2{x}\over (1-x)^2}dx=\sum_{n=0}^{\infty}(1+n)\int_{0}^{1}x^n\ln^2{x}dx\tag3$$

Standard integral $$\int_{0}^{1}x^n\ln^2{x}dx={2\over (1+n)^3}$$

$$J=\sum_{n=0}^{\infty}(1+n)\cdot{2\over (1+n)^3}={\pi^2\over 3}\tag4$$

Substitute $(4)$ into $(2)$

$$I=1\tag5$$

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Hint. Assume $a\ge0,\,k=0,1,2\ldots$. One may consider $$ I_k(a):=\int_{0}^{1}\left(a\cdot x^k+{\ln{x}\over 1-x}\right)^2dx. $$ Then one gets $$ \begin{align} I_k'(a)&=2\int_{0}^{1} x^k\left(a\cdot x^k+{\ln{x}\over 1-x}\right)dx \\\\&=\frac{2 a}{2 k+1}+2\int_{0}^{1} {x^k\ln{x}\over 1-x}dx \\\\&=\frac{2 a}{2 k+1}+2\sum_{j=1}^k\frac1{j^2}-\frac{\pi^2}3. \end{align} $$ Integrating with respect to $a$ gives $$ I_k(a)=\frac{a^2}{2 k+1}+\left(2\sum_{j=1}^k\frac1{j^2}-\frac{\pi^2}3 \right)a+C $$ where $C$ depends only on $k$. Putting $a:=0$ leads to $\displaystyle C=\int_{0}^{1}\left({\ln{x}\over 1-x}\right)^2dx=\frac{\pi^2}3 $ then

$$ \int_{0}^{1}\left(a\cdot x^k+{\ln{x}\over 1-x}\right)^2dx=\frac{a^2}{2 k+1}+\left(2\sum_{j=1}^k\frac1{j^2}-\frac{\pi^2}3 \right)a+\frac{\pi^2}3 $$

from which you deduce the sought integral by putting $a:=1$.

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$$\int_{0}^{1}\left(x^k+{\ln{x}\over 1-x}\right)^2dx=\int_{0}^{1}x^{2k}dx+2\int_{0}^{1}x^k {\ln{x}\over 1-x}dx+\int_{0}^{1}{\ln^2{x}\over (1-x)^2}dx$$

It's obvious that:

$$\int_{0}^{1}x^{2k}dx=\frac{1}{2k+1}$$

And the last integral is already solved by you.

The only integral which can cause you trouble is:

$$\int_{0}^{1}x^k {\ln{x}\over 1-x}dx$$

We can try to solve this integral the following way:

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

$$a=k+l$$

$$\int_0^1 x^k \ln x \sum_{l=0}^{\infty} x^l dx = -\sum_{l=0}^{\infty}\frac{1}{(k+l+1)^2}$$

$$\int_0^1 x^k \frac{\ln x}{1-x}dx=-\psi_1(k+1)=\frac{1}{k^2}-\psi_1(k)= \\ = \frac{1}{k^2}+\frac{1}{(k-1)^2}-\psi_1(k-1)=\dots$$

And:

$$\psi_1(1)=\frac{\pi^2}{6}$$

I don't think there is any prettier expression for general $k$.


As for why you are getting rational numbers - because the third integral is equal to $\pi^2/3$, while the second can be represented as:

$$2\int_{0}^{1}x^k {\ln{x}\over 1-x}dx=-\frac{\pi^2}{3}+2\sum_{m=1}^k \frac{1}{m^2}$$

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\begin{align*} \int_{0}^{1}\left ( x^k + \frac{\ln x}{1-x} \right )^2 \, {\rm d}x&= \int_{0}^{1}x^{2k} \, {\rm d}x + 2\int_{0}^{1}\frac{x^k \ln x} {1-x} \, {\rm d}x + \int_{0}^{1}\frac{\ln^2 x}{(1-x)^2} \, {\rm d}x \\ &= \frac{1}{2k+1} + 2\int_{0}^{1}\frac{x^k \ln x} {1-x} \, {\rm d}x + \int_{0}^{1}\frac{\ln^2 x}{(1-x)^2} \, {\rm d}x \end{align*}

Now the second integral is easy.

\begin{align*} \int_{0}^{1}\frac{x^k \ln x}{1-x} \, {\rm d}x &=\int_{0}^{1} x^k \ln x \sum_{n=0}^{\infty} x^n \, {\rm d}x \\ &= \sum_{n=0}^{\infty} \int_{0}^{1}x^{n+k} \ln x \, {\rm d}x\\ &= -\sum_{n=0}^{\infty} \frac{1}{\left ( n+k+1 \right )^2}\\ &= -\psi^{(0)} (k+1) \\ &= -\zeta(2) + \mathcal{H}^{(2)}_k \end{align*}

And the last one using the $\displaystyle \sum_{n=0}^{\infty}x^n =\frac{1}{1-x}$ and differentiating once we have that:

\begin{align*} \int_{0}^{1}\frac{\ln^2 x}{\left ( 1-x \right )^2} &=\int_{0}^{1} \ln^2 x \sum_{n=1}^{\infty} n x^{n-1} \, {\rm d}x \\ &=\sum_{n=1}^{\infty} n \int_{0}^{1} x^{n-1} \ln^2 x \, {\rm d}x \\ &= 2\sum_{n=1}^{\infty} \frac{n}{n^3}\\ &=\frac{\pi^2}{3} \end{align*}

Thus:

$$\int_{0}^{1}\left ( x^k + \frac{\ln x}{1-x} \right )^2 \, {\rm d}x = \frac{1}{2k+1} +2 \mathcal{H}_k^{(2)}$$

Here $\displaystyle \mathcal{H}_k^{(2)}=\sum_{n=1}^{k} \frac{1}{n^2}$

Woops. I came third. Never mind, since I wrote all this I leave the message.

Edit: Fixed error.

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