Evaluating an indefinite integral I need help evaluating the following indefinite integral explicitly
$$\int \frac{1}{1+t^{2^{-n}}} dt$$
I would appreciate any help

$\frac{1}{1+t^{\frac{1}{2^n}}} = \frac{1}{t^{\frac{1}{2^n}}}\frac{1}{1+\frac{1}{t^{\frac{1}{2^n}}}} = \frac{1}{t^{\frac{1}{2^n}}} \sum_{k=0}^\infty (-1)^k ( \frac{1}{t^{\frac{1}{2^n}}})^k = \sum_{k=0}^\infty (-1)^k  \frac{1}{t^{{\frac{1}{2^n}}(k+1)}}$, and this convergence is uniform for $t \ge 2$.
Hence
\begin{eqnarray}
\int_2^x \frac{1}{1+t^{\frac{1}{2^n}}} dt &=& \sum_{k=0}^\infty (-1)^k \int_2^x \frac{1}{t^{{\frac{1}{2^n}}(k+1)}} dt \\
&=&  \sum_{k=0,\, $k \ne 2^n-1}^\infty (-1)^k \frac{1}{1-{\frac{1}{2^n}}(k+1)} ( \frac{1}{x^{{\frac{1}{2^n}}(k+1)-1}}  - \frac{1}{2^{{\frac{1}{2^n}}(k+1)-1}} ) + (-1)^{2^n-1} \ln \frac{x}{2}\\
&=& \sum_{k=0,\, $k \ne 2^n-1}^\infty (-1)^k \frac{1}{1-{\frac{1}{2^n}}(k+1)} ( \frac{x}{x^{{\frac{1}{2^n}}(k+1)}}  - \frac{2}{2^{{\frac{1}{2^n}}(k+1)}} )  + (-1)^{2^n-1} \ln \frac{x}{2}
\end{eqnarray}
