It's too Obvious!

The substitution $x = e^{-y}$ gives $I(a) \; = \; \int_0^1 \frac{x^a-1}{(x^a+1)\ln x}\,dx \; = \; \int_0^\infty \frac{(1 - e^{-ay})e^{-y}}{(1 + e^{-ay})y}\,dy \; =\; \int_0^\infty \frac{e^{-y}}{1 + e^{-ay}}\int_0^a e^{-uy}\,du\,dy \; = \; \int_0^a \int_0^\infty \frac{e^{-(u+1)y}}{1 + e^{-ay}}\,dy\,du$ for any $a > 0$ . For any $N \in \mathbb{N}$ we define $F_N(u,y) \; = \; \sum_{n=0}^{2N-1} (-1)^n e^{-(u+1+na)y} \hspace{2cm} 0 \le u \le a \,,\, y > 0 \; = \; \frac{e^{-(u+1)y}\big(1 - e^{-2Nay}\big)}{1 + e^{-ay}}$ then $0 \; \le \; \frac{e^{-(u+1)y}}{1 + e^{-ay}} - F_N(u,y) \; = \; \frac{e^{-(u+1+2Na)y}}{1 + e^{-ay}} \; \le \; e^{-2Nay} \hspace{2cm} 0 \le u \le a\,,\,y > 0$ so that $\left| \int_0^a\int_0^\infty \frac{e^{-(u+1)y}}{1 + e^{-ay}} \,dy\,du - \int_0^a \int_0^\infty F_N(u,y)\,dy\,du\right| \; \le \; \frac{1}{2N}$ and hence $\begin{aligned} I(a) & = \; \int_0^a\int_0^\infty \frac{e^{-(u+1)y}}{1 + e^{-ay}} \,dy\,du \;=\; \lim_{N \to \infty}\int_0^a \int_0^\infty F_N(u,y)\,dy\,du \; = \; \lim_{N \to \infty}\sum_{n=0}^{2N-1}(-1)^n \ln\left(\frac{1 + (n+1)a}{1 + na}\right) \\ & = \; \lim_{N \to \infty} \left[\sum_{n=0}^{N-1} \left(\ln\left(\frac{1 + (2n+1)a}{1 + 2na}\right) - \ln\left(\frac{1 + (2n+2)a}{1 + (2n+1)a}\right)\right)\right] \\ & = \; \lim_{N \to \infty}\ln\left(\prod_{n=0}^{N-1} \frac{\big(1 + (2n+1)a\big)^2}{\big(1 + 2na\big)\big(1 + (2n+2)a\big)}\right) \; = \; \lim_{N \to \infty}\ln\left(\prod_{n=0}^{N-1} \frac{\big(n+\frac12(1+a^{-1})\big)^2}{\big(n+\frac12a^{-1}\big)\big(n+1+\frac12a^{-1}\big)}\right) \\ & = \; \lim_{N \to \infty}\ln\left(\frac{G_N\big(\frac12a^{-1}\big)G_N\big(1 + \frac12a^{-1}\big)}{G_N\big(\frac12(1+a^{-1})\big)^2}\right) \end{aligned}$ where $G_N(z) \; = \; \frac{N^z (N-1)!}{z(z+1)(z+2)\cdots (z+N-1)}$ It is well-known that $\lim_{N \to \infty}G_N(z) \; = \; \Gamma(z)$ and hence we deduce that $I(a) \; = \; \ln\left(\frac{\Gamma\big(\frac12a^{-1}\big)\Gamma\big(1 + \frac12a^{-1}\big)}{\Gamma\big(\frac12(1+a^{-1})\big)^2}\right) \hspace{2cm} a > 0$ In this case we have $a = \boxed{1729}$ .

From here you can see that a=1729

Given the title, many people may have just guessed. I did little more, just an educated guess.

Playing around for low values of $a$ in Wolfram Alpha with the function $f(a)=\int_0^1 \frac{x^a-1}{(x^a+1)\ln(x)}dx$ I got $f(1)=\ln(\frac{\pi}{2})$ $f(2)=2\ln \left(\frac{2\Gamma(\frac{5}{4})}{\Gamma(\frac{3}{4})}\right)=\ln\left(\frac{4\Gamma^2(\frac{5}{4})}{\Gamma^2(\frac{3}{4})}\right)\approx 0.783189$ $f(3)\approx 1.03541$ $f(4)=\ln(8)-2\ln(\Gamma(\frac{5}{8}))+2\ln(\Gamma(\frac{9}{8}))=\ln \left(\frac{8 \Gamma^2(\frac{9}{8})}{\Gamma^2(\frac{5}{8})} \right) \approx 1.23774$

WA only gave a closed form for a = 1, 2 or 4, but (using identities like $\Gamma(x+1)=x\Gamma(x)$ ) enough to suggest and numerically confirm a pattern: it seemed that $\int_0^1 \frac{x^a-1}{(x^a+1)\ln(x)}dx =\ln \left( \frac{\Gamma(\frac{1}{2a}+1)\Gamma(\frac{1}{2a})}{\Gamma^2(\frac{1}{2}+\frac{1}{2a})}\right)$