Evaluating $\int_2^4\frac{\arctan x}{\log x}dx$

Question

I've been trying to improve my integration techniques, and I came up with the following integral which I thought I could solve with Feynman Integration
$$I=\int_2^4\frac{\arctan x}{\log x}dx$$
Here's what I've tried so far:
$$F(a)=\int_2^4\frac{\arctan(x^a)}{\log x}dx$$
$$F'(a)=\int_2^4\frac{\partial}{\partial a}\frac{\arctan(x^a)}{\log x}dx$$
$$F'(a)=\int_2^4\frac{x^a}{x^{2a}+1}dx$$
$x=\tan u$: $$F'(a)=\int_{\arctan2}^{\arctan4}\frac{\tan^au\sec^2u}{\tan^{2a}u+1}du$$
$$F'(a)=\int_{\arctan2}^{\arctan4}\frac{\frac{\sin^au}{\cos^{a+2}u}}{\frac{\sin^{2a}u}{\cos^{2a}u}+1}du$$
$$F'(a)=\int_{\arctan2}^{\arctan4}\frac{\sin^au\cos^{2a}u}{\cos^{a+2}u(\sin^{2a}u+\cos^{2a}u)}du$$
$$F'(a)=\int_{\arctan2}^{\arctan4}\frac{\sin^au\cos^{a-2}u}{\sin^{2a}u+\cos^{2a}u}du$$
Which I don't know how to handle. How do I integrate this? Thanks.

Answer 1

NOTE: The following is a partial but incomplete answer.

---

Consider $$F(a) = \int_2^4 \frac{\arctan(x)}{\ln(x)}x^a\,dx.$$ Then $$\frac{d}{da}F(a) = \int_2^4 \frac{\partial}{\partial a} \frac{\arctan(x)}{\ln(x)}x^a\,dx = \int_2^4 \arctan(x) x^{a}\,dx .$$ Note that $$\int \arctan(x) x^{a}\,dx = \frac{\arctan(x) x^{a+1}}{a+1} - \frac{1}{a+1}\int \frac{x^{a+1}}{1+x^2}\,dx.$$ From Wolfram alpha, we have $$\int \frac{x^{a+1}}{1+x^2}\,dx = \frac{1}{2} i^{-a} \left( B(-4; \frac{a}{2}+1,0) - B(-16; \frac{a}{2}+1,0) \right)$$ where $B(z;a,b)$ is the incomplete Beta function. Then we have $$\frac{d}{da}F(a) = \frac{\arctan(4)4^{a+1} - \arctan(2)2^{a+1}}{a+1} - \frac{i^{-a}}{2(a+1)}\left( B(-4; \frac{a}{2}+1,0) - B(-16; \frac{a}{2}+1,0) \right).$$