$\displaystyle \int_{3}^{4} \frac{x^{\frac{1}{7}}}{(7-x)^{\frac{1}{7}} + x^{\frac{1}{7}}} \ dx $
It's extremely simple if you spot the trick (which I didn't).
Let $\displaystyle I_1=\int_3^4\frac{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}\,dx=4-3=1$.
Now,
$\displaystyle I_2=\int_3^4\frac{x^{\frac{1}{7}}}{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}\,dx$
Let $\displaystyle u=7-x\implies -\,du=\,dx$.
So $\displaystyle I_2=\int_3^4\frac{x^{\frac{1}{7}}}{\left(7-x\right)^{\frac{1}{7}}+x^{\frac{1}{7}}}\,dx\xright arrow{u=7-x}{}\int_3^4\frac{\left(7-u\right)^{\frac{1}{7}}}{u^{\frac{1}{7}}+\left(7-u\right)^{\frac{1}{7}}}\,du$ $\displaystyle =\int_3^4\frac{u^{\frac{1}{7}}+\left(7-u\right)^{\frac{1}{7}}}{u^{\frac{1}{7}}+\left(7-u\right)^{\frac{1}{7}}}\,du-\int_3^4\frac{u^{\frac{1}{7}}}{u^{\frac{1}{7}}+\le ft(7-u\right)^{\frac{1}{7}}}\,du$
Therefore,
$\displaystyle 2I_2=I_1=1\implies I_2=\frac{1}{2} $