Why is $\int_{-1}^0x\ln(1+x)dx=3\int_{0}^1x\ln(1+x)dx$?

It sounds like the only rule is that we must avoid explicitly integrating $x\log(1+x)$. If we're equipped with

$$\frac d{dx}\log(1\pm x)=\frac1{x\pm1}$$

then we can convert the logs to definite integrals and apply Fubini's theorem:

$$\begin{align*} I_1 &= \int_{-1}^0 x \log(1+x) \, dx \\ &= - \int_0^1 x \log(1-x) \, dx & x\to-x \\ &= \int_0^1 \int_0^x \cdots \, dy \, dx = \int_0^1 \int_y^1 \frac x{1-y} \, dx \, dy \\ &= \frac12 \int_0^1 (1+y) \, dy \\[2ex] I_2 &= 3 \int_0^1 x \log(1+x) \, dx \\ &= \int_0^1 \int_0^x \cdots \, dy \, dx = \int_0^1 \int_y^1 \frac{3x}{1+y} \, dx \, dy \\ &= \frac32 \int_0^1 (1-y) \, dy \end{align*}$$

By symmetry about $y=\dfrac12$,

$$I_2 - I_1 = \int_0^1 (1-2y)\,dy = 0$$

Note that the following integral vanishes

\begin{align} \int_{0}^1x\ln[(1+x)^3(1-x)]dx=& \ \frac12\int_{0}^1\ln[(1+x)^3(1-x)]d(x^2-1)\\ \overset{ibp}=&\int_0^1 (2x-1)dx=0 \end{align}

We can show that

$$\int_{-1}^{0}x\ln(1+x)dx = -\int_{0}^{1}x\ln(1-x)dx$$

So the limits on the RHS and LHS match the limits in your question. The hard part is finding how

$$-\int_{0}^{1}x\ln(1-x)dx = 3\int_{0}^{1}x\ln(1+x)dx$$

I don't know if this is possible without some sort of evaluation of the integral.