For this question, we need to resolve integrands into partial fractions. The process is rather tedious and time consuming. Therefore I want to minimise the number steps as least as possible. I shall resolve the integral of $$\frac{1}{1+x^{12}}$$ into three partial fractions. If you are interested in it you may try yourself, just read the footnotes or trust me.
\begin{aligned}\int \frac{1}{1+x^{12}} d x&=\frac{1}{3} \left[\int \frac{d x}{1+x^{4}}-\int \frac{x^{4}-2}{1-x^{4}+x^{8}} d x\right]\\&=\frac{1}{3}\left[\underbrace{\int \frac{d x}{1+x^{4}}}_{I_1}-\frac{1}{2} \underbrace{ \int \frac{\sqrt{3} x^{2}-2}{x^{4}-\sqrt{3} x^{2}+1} dx }_{I_2} \displaystyle +\frac{1}{2} \underbrace{\int \frac{\sqrt{3} x^{2}+2}{x^{4}+\sqrt{3} x^{2}+1} d x}_{I_3}\right] \end{aligned}
$$I_1=\displaystyle \int \frac{d x}{1+x^{4}}=\frac{1}{4 \sqrt{2}}\left[2 \tan ^{-1}\left(\frac{x^{2}-1}{2 x}\right)+\ln \left|\frac{x^{2}+\sqrt{2} x+1}{x^{2}-\sqrt{2} x+1}\right|\right]+c_1$$
$$ I_{2} =\int \frac{\sqrt{3} x^{2}-2}{x^{4}-\sqrt{3} x^{2}+1} d x =\int \frac{\sqrt{3}-\frac{2}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-\sqrt{3}} d x$$ $\displaystyle \text {Let } \sqrt{3}-\frac{2}{x^{2}}=A\left(1-\frac{1}{x^{2}}\right)+B\left(1+\frac{1}{x^{2}}\right). $
$\text {Then } A+B=\sqrt{3} \cdots(1) \text { and }-A+B=-2 \cdots(2) $
$(1)-(2) \text { gives } \displaystyle A=\frac{\sqrt{3}+2}{2} \\$ $(1)+(2) \text { gives } \displaystyle B=\frac{\sqrt{3}-2}{2}$
\begin{aligned}I_{2} &=\frac{\sqrt{3}+2}{2} \int \frac{1-\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-\sqrt{3}} dx +\frac{\sqrt{3}-2}{2} \int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-\sqrt{3}} d x \\ &=\frac{\sqrt{3}+2}{2} \int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-(2+\sqrt{3})} +\frac{\sqrt{3}-2}{2} \int \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+(2-\sqrt{3})} \\ & =\frac{\sqrt{3}+2}{4 \sqrt{2+\sqrt{3}}} \ln \left|\frac{x+\frac{1}{x}-\sqrt{2+\sqrt{3}}}{x+\frac{1}{x}+\sqrt{2+\sqrt{3}}}\right|+\frac{\sqrt{3}-2}{2 \sqrt{2-\sqrt{3}}} \tan ^{-1} \frac{x-\frac{1}{x}}{\sqrt{2-\sqrt{3}}} +c_2\\ &=\frac{\sqrt{3}+1}{4 \sqrt{2}} \ln \left| \frac{\sqrt{2} x^{2}-(\sqrt{3}+1) x+\sqrt{2}}{\sqrt{2} x^{2}+(\sqrt{3}+1) x+\sqrt{2}}\right|\quad -\frac{\sqrt{3}-1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{\sqrt{2} \left(x^{2}-1\right)}{(\sqrt{3}-1) x}\right)+c_2 \end{aligned}
Similarly,
$$\quad I_{3}=-\frac{\sqrt{3}-1}{4 \sqrt{2}} \ln \left|\frac{\sqrt{2} x^{2}-(\sqrt{3}-1) x+\sqrt{2}}{\sqrt{2} x^{2}+(\sqrt{3}-1) x+\sqrt{2}} \right|+\frac{\sqrt{3}+1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{\sqrt{2}\left(x^{2}-1\right)}{(\sqrt{3}+1) x}\right)+c_3$$
We can now conclude that
\begin{aligned} I &=\frac{1}{3}\left(I_{1}-\frac{1}{2} I_{2}+\frac{1}{2} I_{3}\right)\\ \displaystyle &=\frac{1}{24 \sqrt{2}} \Big[2 \ln \left|\frac{x^{2}+\sqrt{2} x+1}{x^{2}-\sqrt{2} x+1}\right|+4 \tan ^{-1}\left(\frac{x^{2}-1}{2 x}\right)\\ \displaystyle &\quad +(\sqrt{3}+1) \ln \left|\frac{\sqrt{2} x^{2}+(\sqrt{3}+1) x+\sqrt{2}}{\sqrt{2} x^{2}-(\sqrt{3}+1) x+\sqrt{2}}\right|+2(\sqrt{3}-1) \tan ^{-1}\left(\frac{\sqrt{2}\left(x^{2}-1\right)}{(\sqrt{3}-1) x}\right)\\ \displaystyle &\quad +(\sqrt{3}-1) \ln \left|\frac{\sqrt{2} x^{2}+(\sqrt{3}-1) x+\sqrt{2}}{\sqrt{2} x^{2}-(\sqrt{3}-1) x+\sqrt{2}}\right|+2(\sqrt{3}+1) \tan ^{-1}\left(\frac{\sqrt{2}\left(x^{2}-1\right)}{(\sqrt{3}+1) x}\right)\Big]+C \end{aligned}
Is there other any simpler method?
For completeness and checking, I also evaluate
$\displaystyle \quad \int_0^1 \frac{1}{1+x^{12}}dx$ $\displaystyle =\frac{1}{24 \sqrt{2}}\left[2 \ln \left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right)+(1+\sqrt{3}) \ln \left(\frac{2 \sqrt{2}+\sqrt{3}+1}{2 \sqrt{2}-\sqrt{3}-1}\right)\right.\\$ $\displaystyle \left.+(\sqrt{3}-1) \ln \left(\frac{2 \sqrt{2}+\sqrt{3}-1}{2 \sqrt{2}-\sqrt{3}+1}\right)+2(1+\sqrt{3}) \pi\right] \\$ $ \doteq 0.94747$ (checked by Wolframalpha)
