1. The problem is to evaluate the integral $$\int_0^1 \frac{\arctan(x) \cdot (\log x)^3}{1+x} \, dx.$$\n\n2. This integral involves the arctangent function, logarithms, and a rational function. Such integrals often require series expansions or special function representations.\n\n3. Recall the series expansion for arctan(x): $$\arctan(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}.$$\n\n4. Also, note that $$\frac{1}{1+x} = \sum_{m=0}^\infty (-1)^m x^m$$ for $|x|<1$.\n\n5. Substitute these into the integral: $$\int_0^1 \arctan(x) \frac{(\log x)^3}{1+x} dx = \int_0^1 \left( \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1} \right) (\log x)^3 \left( \sum_{m=0}^\infty (-1)^m x^m \right) dx.$$\n\n6. Interchange summation and integration (justified by uniform convergence): $$= \sum_{n=0}^\infty \sum_{m=0}^\infty (-1)^{n+m} \frac{1}{2n+1} \int_0^1 x^{2n+1+m} (\log x)^3 dx.$$\n\n7. The integral $$\int_0^1 x^a (\log x)^3 dx$$ for $a > -1$ is known: $$\int_0^1 x^a (\log x)^3 dx = -\frac{6}{(a+1)^4}.$$\n\n8. Applying this, we get $$\int_0^1 x^{2n+1+m} (\log x)^3 dx = -\frac{6}{(2n + m + 2)^4}.$$\n\n9. Substitute back: $$\sum_{n=0}^\infty \sum_{m=0}^\infty (-1)^{n+m} \frac{1}{2n+1} \left(-\frac{6}{(2n + m + 2)^4}\right) = -6 \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{n+m}}{2n+1} \frac{1}{(2n + m + 2)^4}.$$\n\n10. This double series converges and represents the exact value of the integral.\n\nFinal answer: $$\int_0^1 \frac{\arctan(x) (\log x)^3}{1+x} dx = -6 \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(-1)^{n+m}}{2n+1} \frac{1}{(2n + m + 2)^4}.$$