From 0 to 1

Let $x^2$ be $\frac{du}{dx}$ and $tan^{-1} x$ be $v$ . Then $u=\frac{1}{3}$ $x^3$ while $\frac{dv}{dx} =\frac{1}{1+ x^2}$ . Now we get $(\frac{1}{3}x^3)tan^{-1}x$ minus the antiderivative of $(\frac{1}{3}x^3) \frac{1}{1+x^2}$ . Substituting 1 and 0 into $(\frac{1}{3}$ $x^3) (tan^{-1}x)$ gives us $\frac{1}{3}(\frac{\pi}{4})-0 = \frac{\pi}{12}$

Now rewrite $\frac{1}{3} \frac{x^3}{1+x^2}$ as $\frac{1}{3} (x-\frac{x}{1+x^2})$ . Integrating this gives us $\frac{1}{3} [\frac{1}{2}(x^2) - \frac{1}{2} ln (1+ x^2)] = \frac{1}{6} [x^2 - ln (1+ x^2)]$ . Substituting 1 and 0 into this gives us $\frac{1}{6}- \frac{ln 2}{6}-0$ .

It follows that the answer is $\frac{\pi}{12}$ - $\frac{1}{6}$ + $\frac{ln 2}{6}$ and we have $12 + 1 + 6 + 2 + 6 =\boxed {27}$ .