May be too lengthy 2

Using the Maclaurin series expansion for $\tan^{-1}x$ we define $f(a) \; = \; \int_0^1 x^a \tan^{-1}x\,dx \; = \; \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+a+2)} \;, \qquad a \ge -1\;,$ so that $f^{(j)}(a) \; = \; (-1)^j j! \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+a+2)^{j+1}} \;, \qquad j \ge 0 \;,$ (all these series converge absolutely and uniformly for $a \ge -1$ , so the term-by-term differentiation is justified). Thus $f''(0) \; = \; 2\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)^3} \qquad f'''(0) \; =\; -6\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)^4}$ and so $\begin{array}{rcl} \displaystyle \int_0^1 \tan^{-1}x\big[(\ln x)^3 + 3(\ln x)^2\big]\,dx & = & f'''(0) + 3f''(0) \\ & = & \displaystyle 6\sum_{n=0}^\infty \left[ \frac{(-1)^n}{(2n+1)(2n+2)^3} - \frac{(-1)^n}{(2n+1)(2n+2)^4}\right] \\ & = & \displaystyle 6\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)^4}\big[(2n+2)-1\big] \\ & = & \displaystyle 6\sum_{n=0}^\infty \frac{(-1)^n}{(2n+2)^4} \; = \; \frac38 \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^4} \\ & = & \displaystyle \frac38\left[ \sum_{n=1}^\infty \frac{1}{n^4} - 2\sum_{n=1}^\infty \frac{1}{(2n)^4}\right] \\ & = &\displaystyle \tfrac38\left[\zeta(4) - \tfrac18\zeta(4)\right] \; = \; \tfrac{21}{64}\zeta(4) \\ & = & \frac{7}{1920}\pi^4 \;, \end{array}$ making the answer $7 + 1920 + 4 \,=\, \boxed{1931}$ .