Integrating $\int_0^1 \frac{\ln(1+x)\ln^3 x}{1+x}\,dx$ with restricted techniques

How does one calculate these four following integrals?

$$ \int_0^1\frac{\ln(1\pm\varepsilon x)\ln(x)^3}{1\pm \varepsilon x}\,dx,\;\forall\varepsilon\in\{-1,1\}. $$ CONTEXT: Our teacher asks us to to calculate these four integrals using only changes of variables, integrations by parts and the following known result: $$\int_0^1 \frac{\ln^n(x)}{1-x} \; dx=(-1)^n n!\zeta(n+1),\quad \int_0^1 \frac{\ln^n(x)}{1+x} \; dx=(-1/2)^n (-1 + 2^n) \Gamma(1 + n) \zeta (1 + n)$$ without using complex analysis, series, differentiation under the integral sign, double integrals or special functions.

For calculate $ U =\int_0^1 \frac{\ln(1+x)\ln^3 x}{1-x}\,dx\\$ By IBP

$ U =\left[\left(\int_0^x \frac{\ln^3t}{1-t}\,dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x\frac{\ln^3t}{1-t}\,dt\right)\,dx\\ =-6\zeta(4)\ln 2+\int_0^1\int_0^1\left(\frac{\ln^3(tx)}{(1+t)(1+x)}-\frac{\ln^3(tx)}{(1+t)(1-tx)}\right)\,dt\,dx\\ =-6\zeta(4)\ln 2+6\left(\int_0^1\frac{\ln^2 t}{1+t}\,dt\right)\left(\int_0^1\frac{\ln x}{1+x}\,dx\right)+2\left(\int_0^1\frac{\ln^3 t}{1+t}\,dt\right)\left(\int_0^1\frac{1}{1+x}\,dx\right)-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\$ $=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\ \overset{\text{IBP}}=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\left[\ln\left(\frac {t}{1+t}\right)\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\right]_0^1+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\ =-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\ =-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+24\zeta(5)-U\\ U =\boxed{-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)}$

Precisely, i can't see how to calculate $V=\int_0^1 \frac{\ln(1+x)\ln^3 x}{1+x}\,dx\\$ edit I am also interested in how to calculate the two other integrals

\begin{align*} K&=\int_0^1 \frac{\ln(1-x)\ln^3 x}{1-x}\,dx,C=\int_0^1 \frac{\ln^3 t}{1-t}\,dt\\ K&=\left[\left(\int_0^x \frac{\ln^3 t}{1-t}\,dt-C\right)\ln(1-x)\right]_0^1+\int_0^1 \frac{1}{1-x}\left(\int_0^x \frac{\ln^3 t}{1-t}\,dt-C\right)\,dx\\ &=\int_0^1 \left(\left(\int_0^1 \frac{x\ln^3(tx)}{(1-x)(1-tx)}\,dt\right)-\frac{C}{1-x}\right)\,dx\\ &=\int_0^1 \left(\left(\int_0^1 \frac{\ln^3(tx)}{(1-t)(1-x)}\,dt-\int_0^1 \frac{\ln^3(tx)}{(1-t)(1-tx)}\,dt\right)-\frac{C}{1-x}\right)\,dx\\ &=6\left(\int_0^1 \frac{\ln^2 t}{1-t}\,dt\right)\left(\int_0^1 \frac{\ln x}{1-x}\,dx\right)+\\ &\int_0^1 \left(\left(\int_0^1 \frac{\ln^3(x)}{(1-t)(1-x)}\,dt-\int_0^1 \frac{\ln^3(tx)}{(1-t)(1-tx)}\,dt\right)+\frac{C}{1-x}-\frac{C}{1-x}\right)\,dx\\ &=6\left(\int_0^1 \frac{\ln^2 t}{1-t}dt\right)\left(\int_0^1 \frac{\ln x}{1-x}dx\right)+\int_0^1 \left(\int_0^1 \frac{\ln^3(x)}{(1-t)(1-x)}dt-\int_0^1 \frac{\ln^3(tx)}{(1-t)(1-tx)}dt\right)dx\\ &=-12\zeta(2)\zeta(3)+\int_0^1 \left(\int_0^1 \frac{\ln^3(x)}{(1-t)(1-x)}dt-\int_0^1 \frac{\ln^3(tx)}{(1-t)(1-tx)}dt\right)dx\\ &=-12\zeta(2)\zeta(3)+\int_0^1 \left(\frac{C}{1-t}-\frac{1}{t(1-t)}\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)\right)dt\\ 0&\leq A<1\\ K(A)&=-12\zeta(2)\zeta(3)+\int_0^A \left(\frac{C}{1-t}-\frac{1}{t(1-t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\right)\,dt\\ &=-12\zeta(2)\zeta(3)-C\ln(1-A)-\int_0^A \frac{1}{t(1-t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\ &=-12\zeta(2)\zeta(3)-C\ln(1-A)-\left[\ln\left(\frac{t}{1-t}\right)\left(\int_0^t \frac{\ln^3 u}{1-u}du\right)\right]_0^A+\\ &\int_0^A \frac{\ln\left(\frac{t}{1-t}\right)\ln^3 t}{1-t}dt\\ &=-12\zeta(2)\zeta(3)+\ln(1-A)\left(\left(\int_0^A \frac{\ln^3 u}{1-u}du\right)-C\right)-\ln A\left(\int_0^A \frac{\ln^3 u}{1-u}du\right)+\\ &\int_0^A \frac{\ln^4 t}{1-t}\,dt-\int_0^A \frac{\ln(1-t)\ln^3 t}{1-t}\,dt\\ K&=\lim_{A\rightarrow 1}K(A)\\ &=-12\zeta(2)\zeta(3)+\int_0^1 \frac{\ln^4 t}{1-t}\,dt-K\\ &=-12\zeta(2)\zeta(3)+24\zeta(5)-K\\ K&=\boxed{12\zeta(5)-6\zeta(2)\zeta(3)} \end{align*} "Easier", \begin{align*} U&=\int_0^1 \frac{\ln(1+x)\ln^3 x}{1-x}\,dx\\ U&\overset{\text{IBP}}=\left[\left(\int_0^x \frac{\ln^3 t}{1-t}\,dt\right)\ln(1+x)\right]_0^1-\int_0^1 \frac{1}{1+x}\left(\int_0^x\frac{\ln^3 t}{1-t}\,dt\right)\,dx\\ &=-6\zeta(4)\ln 2+\int_0^1\int_0^1 \left(\frac{\ln^3(tx)}{(1+t)(1+x)}-\frac{\ln^3(tx)}{(1+t)(1-tx)}\right)dt dx\\ &=-6\zeta(4)\ln 2+6\left(\int_0^1\frac{\ln^2 t}{1+t}dt\right)\left(\int_0^1\frac{\ln x}{1+x}dx\right)+2\left(\int_0^1\frac{\ln^3 t}{1+t}dt\right)\left(\int_0^1\frac{1}{1+x}dx\right)-\\ &\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\ &=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\int_0^1 \frac{1}{t(1+t)}\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\,dt\\ &\overset{\text{IBP}}=-\frac{33}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)-\left[\ln\left(\frac{t}{1+t}\right)\left(\int_0^t \frac{\ln^3 u}{1-u}\,du\right)\right]_0^1+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\ &=-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+\int_0^1 \frac{\ln\left(\frac{t}{1+t}\right)\ln^3 t}{1-t}\,dt\\ &=-\frac{45}{2}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)+24\zeta(5)-U\\ U&=\boxed{-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)} \end{align*}

NB: The same method doesn't work fine for the two others integrals.

I assume that, \begin{align} \int_0^1 \frac{\ln^4 x}{1+x}\,dx&=\frac{45}{2}\zeta(5), \int_0^1 \frac{\ln^4 x}{1-x}\,dx=24\zeta(5)\\ \int_0^1 \frac{\ln^3 x}{1-x}\,dx&=-6\zeta(4), \int_0^1 \frac{\ln^3 x}{1+x}\,dx=-\frac{21}{4}\zeta(4)\\ \int_0^1 \frac{\ln^2 x}{1-x}\,dx&=2\zeta(3), \int_0^1 \frac{\ln^2 x}{1+x}\,dx=\frac{3}{2}\zeta(3)\\ \int_0^1 \frac{\ln x}{1-x}\,dx&=-\zeta(2), \int_0^1 \frac{\ln x}{1+x}\,dx=-\frac{1}{2}\zeta(2)\\ \end{align}

Addendum

Computation of the two other integrals. \begin{align*} K_1&=K,K_2=\int_0^1 \frac{\ln(1+x)\ln^3 x}{1+x}\,dx\\V&=\int_0^1 \frac{\ln(1-x)\ln^3 x}{1+x}\,dx\\ K_1+U-K_2-V&=\int_0^1 \frac{\ln(1-x^2)\ln^3 x}{1-x}\,dx-\int_0^1 \frac{\ln(1-x^2)\ln^3 x}{1+x}\,dx\\ &=\int_0^1 \frac{2x\ln(1-x^2)\ln^3 x}{1-x^2}\,dx\\ &\overset{y=x^2}=\frac{1}{8}K_1\\ V&=\frac{7}{8}K_1+U-K_2\\ \end{align*}

\begin{align*} C_1&=\int_0^{\frac{1}{2}}\frac{\ln^4 x}{1-x}\,dx,K_3=\int_0^{\frac{1}{2}}\frac{\ln^3 x\ln(1-x)}{1-x}\,dx\\ \int_0^1 \frac{\ln^4\left(\frac{x}{1+x}\right)}{1+x}\,dx&=\int_0^1 \frac{\ln^4\left(1+x\right)}{1+x}\,dx-4\int_0^1 \frac{\ln^3\left(1+x\right)\ln x}{1+x}\,dx+\\&6\int_0^1\frac{\ln^2\left(1+x\right)\ln^2 x}{1+x}\,dx-4\int_0^1 \frac{\ln\left(1+x\right)\ln^3 x}{1+x}\,dx+\int_0^1 \frac{\ln^4 x}{1+x}\,dx\\ &=\frac{1}{5}\ln^5 2-\Big[\ln^4(1+x)\ln x\Big]_0^1+\int_0^1 \frac{\ln^4(1+x)}{x}\,dx+\\ &2\Big[\ln^3(1+x)\ln^2 x\Big]_0^1-4\int_0^1 \frac{\ln^3(1+x)\ln x}{x}\,dx-4K_1+\frac{45\zeta(5)}{2}\\ &=\frac{1}{5}\ln^5 2-4\int_0^1 \frac{\ln^3(1+x)\ln x}{x}\,dx+\int_0^1 \frac{\ln^4(1+x)}{x}\,dx-4K_2+\frac{45\zeta(5)}{2}\\ &\overset{y=\frac{1}{1+x}}=\frac{1}{5}\ln^5 2-4\int_{\frac{1}{2}}^1\frac{\ln^3 y\Big(\ln y-\ln(1-y)\Big)}{y(1-y)}dy+\int_{\frac{1}{2}}^1\frac{\ln^4 y}{y(1-y)}dy-\\&4K_2+\frac{45\zeta(5)}{2}\\ &=\frac{1}{5}\ln^5 2-3\int_{\frac{1}{2}}^1\frac{\ln^4 x}{x}\,dx-3\int_{\frac{1}{2}}^1\frac{\ln^4 x}{1-x}\,dx+4\int_{\frac{1}{2}}^1\frac{\ln^3 x\ln(1-x)}{x}\,dx+\\ &4\int_{\frac{1}{2}}^1\frac{\ln^3 x\ln(1-x)}{1-x}\,dx-4K_2+\frac{45}{2}\zeta(5)\\ &=-\frac{2}{5}\ln^5 2-3\Big(24\zeta(5)-C_1\Big)+4\int_{\frac{1}{2}}^1\frac{\ln^3 x\ln(1-x)}{x}\,dx+\\&4\Big(K_1-K3\Big)-4K_2+\frac{45}{2}\zeta(5)\\ &\overset{\text{IBP}}=-\frac{2}{5}\ln^5 2-3\Big(24\zeta(5)-C_1\Big)+\Big[\ln^4 x\ln(1-x)\Big]_{\frac{1}{2}}^1+\\&\int_{\frac{1}{2}}^1\frac{\ln^4 x}{1-x}\,dx+4\Big(K_1-K3\Big)-4K_2+\frac{45\zeta(5)}{2}\\ &=\frac{3}{5}\ln^5 2-2\Big(24\zeta(5)-C_1\Big)+4\Big(K_1-K3\Big)-4K_2+\frac{45\zeta(5)}{2}\\ \end{align*} On the other hand,

\begin{align*} \int_0^1 \frac{\ln^4\left(\frac{x}{1+x}\right)}{1+x}\,dx&\overset{y=\frac{x}{1+x}}=C_1 \end{align*} Therefore, \begin{align*} C_1&=\frac{3}{5}\ln^5 2-2\Big(24\zeta(5)-C_1\Big)+4\Big(K_1-K3\Big)-4K_2+\frac{45}{2}\zeta(5)\\ K_2+K_3&=\frac{3}{20}\ln^5 2+\frac{1}{4}C_1+K_1-\frac{51}{8}\zeta(5) \end{align*} Moreover, \begin{align*} K_2&\overset{y=\frac{1}{1+x}}=\int_{\frac{1}{2}}^1\frac{\ln y\Big(\ln y-\ln(1+y)\Big)^3}{y}\,dy\\ &=\int_{\frac{1}{2}}^1\left(\frac{\ln^4 y}{y}-\frac{3\ln(1-y)\ln^3 y}{y}+\frac{3\ln^2(1-y)\ln^2 y}{y}-\frac{\ln^3(1-y)\ln y}{y}\right)\,dy\\ &\overset{\text{IBP}}=\frac{1}{5}\ln^5 2-\frac{3}{4}\Big[\ln^4 y\ln(1-y)\Big]_{\frac{1}{2}}^1-\frac{3}{4}\int_{\frac{1}{2}}^1 \frac{\ln^4 y}{1-y}\,dy+\Big[\ln^3 y\ln^2(1-y)\Big]_{\frac{1}{2}}^1+\\ &2\int_{\frac{1}{2}}^1 \frac{\ln^3 y\ln(1-y)}{1-y}\,dy-\int_{\frac{1}{2}}^1 \frac{\ln^3(1-y)\ln y}{y}\,dy\\ &=\frac{9}{20}\ln^5 2-\frac{3}{4}\Big(24\zeta(5)-C1\Big)+2\Big(K_1-K3\Big)-\int_{\frac{1}{2}}^1 \frac{\ln^3(1-y)\ln y}{y}\,dy\\ &\overset{z=1-y}=\frac{9}{20}\ln^5 2-\frac{3}{4}\Big(24\zeta(5)-C1\Big)+2\Big(K_1-K3\Big)-K3\\ K_2+3K_3&=\frac{9}{20}\ln^5 2-18\zeta(5)+\frac{3}{4}C_1+2K_1\\ \end{align*} Therefore, \begin{align*} K_2&=\boxed{\frac{1}{2}K_1-\frac{9}{16}\zeta(5)}\\ &=\frac{1}{2}\Big(12\zeta(5)-6\zeta(2)\zeta(3)\Big)-\frac{9}{16}\zeta(5)\\ &=\boxed{\frac{87}{16}\zeta(5)-3\zeta(2)\zeta(3)}\\ V&=\frac{7}{8}K_1+\left(-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)+12\zeta(5)\right)-\left(\frac{1}{2}K_1-\frac{9}{16}\zeta(5)\right)\\ &=\frac{3}{8}K_1+\frac{201}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)\\ &=\frac{3}{8}\Big(12\zeta(5)-6\zeta(2)\zeta(3)\Big)\Big)+\frac{201}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{4}\zeta(2)\zeta(3)\\ &=\boxed{\frac{273}{16}\zeta(5)-\frac{45}{4}\zeta(4)\ln 2-\frac{9}{2}\zeta(2)\zeta(3)} \end{align*}

$$\mathcal{I}=\int_0^1\frac{\ln^3x\ln(1+x)}{x(1+x)}dx=\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}dx-\underbrace{\int_1^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}dx}_{x\mapsto 1/x}$$

$$\mathcal{I}=\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}dx+\color{blue}{\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}dx}-\int_0^1\frac{\ln^4x}{1+x}dx$$

By adding $\ \mathcal{I}=\int_0^1\frac{\ln^3x\ln(1+x)}{x(1+x)}dx=\int_0^1\frac{\ln^3x\ln(1+x)}{x}dx-\color{blue}{\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}dx}\ $ to both sides, the blue integral nicely cancels out and we get

$$2\mathcal{I}=\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}dx-\int_0^1\frac{\ln^4x}{1+x}dx+\underbrace{\int_0^1\frac{\ln^3x\ln(1+x)}{x}dx}_{IBP}$$

$$2\mathcal{I}=\underbrace{\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}dx}_{\text{Beta function:}\ 6\zeta(2)\zeta(3)+6\zeta(5)}-\frac54\underbrace{\int_0^1\frac{\ln^4x}{1+x}dx}_{\frac{45}2\zeta(5)}$$

or

$$\mathcal{I}=3\zeta(2)\zeta(3)-\frac{177}{16}\zeta(5)\tag1$$

But

$$\mathcal{I}=\int_0^1\frac{\ln^3x\ln(1+x)}{x}dx-\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}dx$$

$$=-\frac{45}{8}\zeta(5)-\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}dx\tag2$$

Subtarcting (1) and (2) yields

$$\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}dx=\frac{87}{16} \zeta(5)-3\zeta(2)\zeta(3)$$

The integral $\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}dx$ can be calculated without using beta function:

With $\frac1{1+x}=y$ we have

$$\int_0^\infty\frac{\ln^3x\ln(1+x)}{x(1+x)}\ dx=\int_0^1\frac{\ln^3\left(\frac{x}{1-x}\right)\ln x}{1-x}\ dx$$

$$=\int_0^1\frac{\ln^4x}{1-x}-3\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}+3\underbrace{\int_0^1\frac{\ln^2x\ln^2(1-x)}{1-x}}_{IBP}-\underbrace{\int_0^1\frac{\ln x\ln^3(1-x)}{1-x}\ dx}_{IBP}$$

$$=\int_0^1\frac{\ln^4x}{1-x}-3\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}+2\underbrace{\int_0^1\frac{\ln^3(1-x)\ln x}{x}}_{\large 1-x\to x}-\frac14\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{\large 1-x\to x}$$

$$=\frac34\int_0^1\frac{\ln^4x}{1-x}\ dx-\int_0^1\frac{\ln^3x\ln(1-x)}{1-x}\ dx$$ $$=\frac34(4!\zeta(5))+\sum_{n=1}^\infty H_n\int_0^1 x^n \ln^3x\ dx$$

$$=18\zeta(5)-6\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}$$

$$=18\zeta(5)-6\sum_{n=1}^\infty\frac{H_n}{n^4}+6\zeta(5)$$

$$=18\zeta(5)-6[3\zeta(5)-\zeta(2)\zeta(3)]+6\zeta(5)$$

$$=6\zeta(2)\zeta(3)+6\zeta(5)$$

Your integral can be related to harmonic series:

$$\int_0^1\frac{\ln^3x\ln(1+x)}{1+x}dx=-\sum_{n=1}^\infty (-1)^nH_n\int_0^1 x^n \ln^3xdx$$

$$=6\sum_{n=1}^\infty\frac{(-1)^nH_n}{(n+1)^4}=-6\sum_{n=1}^\infty\frac{(-1)^nH_{n-1}}{n^4}$$ $$=-6\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^4}+6\sum_{n=1}^\infty\frac{(-1)^n}{n^5}$$

$$=-6\sum_{n=1}^\infty\frac{(-1)^nH_{n}}{n^4}-\frac{45}{8}\zeta(5)$$