Not a fun Integration By Parts...

Let $x=u^7,dx=7u^6du$ , then the integral $I$ becomes $I=\displaystyle\int_0^1 7u^6\arctan u du$ Then by integration by parts, we get $I=u^7\arctan u\bigg|^1_0-\displaystyle\int^1_0 \dfrac{u^7}{1+u^2}du=\dfrac{\pi}{4}-I_1$ As $\dfrac{u^7}{1+u^2}=u^5-u^3+u+\dfrac{u}{1+u^2}$ , $I_1=\displaystyle\int^1_0 \left(u^5-u^3+u-\dfrac{u}{1+u^2}\right)du=\dfrac{u^6}{6}-\dfrac{u^4}{4}+\dfrac{u^2}{2}-\dfrac{1}{2}\ln(u^2+1)\bigg|^1_0=\dfrac{5}{12}-\dfrac{\ln 2}{2}$ So therefore $I=\dfrac{\pi}{4}-\left(\dfrac{5}{12}-\dfrac{\ln 2}{2}\right)=\dfrac{\ln 2}{2}+\dfrac{\pi}{4}-\dfrac{5}{12}$ Then $A=2,B=2,C=1,D=4,E=5,F=12,A+B+C+D+E+F=\boxed{26}$

The given integral

$I = \int_{0}^{1} \arctan\left(x^{1/7}\right)dx$

Integrating by parts yields:

$I = \left. x\arctan\left(x^{1/7}\right) \right\vert_{0}^{1} - \frac{1}{7}\int_{0}^{1} \frac{x^{1/7}dx}{1+x^{2/7}}$ $I = \frac{\pi}{4} - I_2$

Where:

$I_2 = \frac{1}{7}\int_{0}^{1} \frac{x^{1/7}dx}{1+x^{2/7}}$

The following substitution:

$x^{1/7} = \tan{p} \implies x = \tan^7(p) \implies dx = 7\sec^2(p)\tan^6(p)dp$

This transforms the integral to:

$I_2 = \int_{0}^{\pi/4} \tan^7(p)dp$

Consider the general integral:

$J_n = \int_{0}^{\pi/4} \tan^n(p)dp \implies \int_{0}^{\pi/4} \tan^{n-2}(p)\left(\sec^2(p)-1\right)dp$ $J_n = \int_{0}^{\pi/4} \tan^{n-2}(p)\sec^2(p) - J_{n-2}$ $J_n = \left.\frac{\tan^{n-1}(p)}{n-1}\right\vert_{0}^{\pi/4} -J_{n-2}$ $\boxed{J_n = \frac{1}{n-1} - J_{n-2}}$

Now, applying this formula recursively:

$I_2 = J_7 = \frac{1}{6} - J_5 \implies I_2= \frac{1}{6} - \frac{1}{4} + \frac{1}{2} - J_1 = \frac{5}{12}-J_1$ $I_2 = \frac{5}{12} - \int_{0}^{\pi/4} \tan(p)dp \implies I_2 = \frac{5}{12} - \frac{\ln{2}}{2}$

Therefore:

$I = \frac{\pi}{4} - I_2 \implies \boxed{I = \frac{\pi}{4} + \frac{\ln{2}}{2} - \frac{5}{12}}$

I have skipped explaining steps which I consider trivial. I will elaborate if requested.

We have $\begin{aligned} \frac1{1+x^2} &= 1-x^2+x^4-\cdots \\ \arctan(x) &= x-\frac13 x^3 + \frac15 x^5 - \cdots \\ \arctan(\sqrt[7]{x}) &= x^{1/7} - \frac13 x^{3/7} + \frac15 x^{5/7} - \cdots \\ \int_0^1 \arctan(\sqrt[7]{x}) \, dx &= \frac78 x^{8/7} - \frac7{10} \frac13 x^{10/7} + \frac7{12} \frac15 x^{12/7} - \cdots \bigg|_0^1 \\ &= \frac7{1 \cdot 8} - \frac7{3 \cdot 10} + \frac7{5 \cdot 12} - \cdots \\ &= \frac11 - \frac18 -\frac13 + \frac1{10} + \frac15 - \frac1{12} - \cdots \end{aligned}$ Now $\frac11 - \frac13 + \frac15 - \cdots = \frac{\pi}4,$ and $\begin{aligned} -\frac18 + \frac1{10} - \frac1{12} + \cdots &= -\frac12 + \frac14 - \frac16 + \left( \frac12 - \frac14 + \frac16 - \frac18 + \frac1{10} - \frac1{12} + \cdots \right) \\ &= -\frac5{12} + \frac12 \left( 1 - \frac12 + \frac13 - \frac14 + \cdots \right) \\ &= -\frac5{12} + \frac12 \ln(2), \end{aligned}$ so the integral is $\frac{\pi}4-\frac5{12} + \frac{\ln(2)}2$ and the answer is $2+2+1+4+5+12 = \fbox{26}.$

(I have avoided difficult questions about the legality of rearranging terms in these conditionally convergent series because I am too lazy to resolve them!)

Observe that $\sqrt[7]{x}$ is invertible. This allows us to safely make the parametrization $y=\arctan t$ and $t=\sqrt[7]{x}$ (or $x=t^7.$ The following is a known result for integrals of parametrized variables: If $y=F(x)$ where $x=f(t)$ and $y=g(t)$ , then $\int_a^bF(x)\ dx=\int_\alpha^\beta g(t)f'(t)$ where $f(\alpha)=a$ and $f(\beta)=b$ .

Setting $f(t)=t^7$ and $g(t)=\arctan t$ yields the following:

$\int_0^17x^6\arctan x\ dx=x^7\arctan x|^1_0-\int_0^1\dfrac{x^7}{x^2+1}\ dx=\dfrac{\pi}{4}-\int_0^1\dfrac{x^7}{x^2+1}\ dx$ This can simply be solved with polynomial long division and a $u$ -substitution. $\int_0^1\dfrac{x^7}{x^2+1}\ dx=\int_0^1\left(x^5-x^3+x-\dfrac{x}{x^2+1}\right)\ dx=\dfrac{1}{6}-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{2}\int_1^2\dfrac{du}{u}=\dfrac{5}{12}-\dfrac{\ln2}{2}$ Thus, the overall integral is equal to $\dfrac{\ln2}{2}+\dfrac{\pi}{4}-\dfrac{5}{12}.$ $\{A,B,C,D,E,F\}=\{2,2,1,4,5,12\},$ so $A+B+C+D+E+F=\boxed{26}.$