Evaluation of $\int\frac{1}{2014^x+2015^x}dx$ Evaluation of $\displaystyle \int\frac{1}{2014^x+2015^x}dx$
$\bf{My\; Try::}$ Let $\displaystyle I = \int\frac{1}{2014^x+2015^x}dx = \int\frac{2014^{-x}}{1+\left(\frac{2015}{2014}\right)^x}dx$
Now How can I solve after that
Help me
Thanks
 A: It seems clear that there is no closed form for the antiderivative. However, some approximations can be made but their validity would need to be seriously checked.
Let us consider $$I=\int \frac{dx}{e^{k_1x}+e^{k_2x}}=\int \frac{e^{-k_1x}}{1+e^{(k_2-k_1)x}}dx$$ and now suppose that we can approximate the denominator by a first order Taylor expansion built at $x=0$ (or at the lower bound of integration). Then $$I = \int \frac{e^{-k_1x}}{1+e^{(k_2-k_1)x}}dx \simeq \int\frac{e^{-k_1x}}{2+(k_2-k_1)x}dx=\frac{e^{-\frac{2 {k_1}}{{k_1}-{k_2}}} \text{Ei}\left(\frac{2
   {k_1}}{{k_1}-{k_2}}-{k_1} x\right)}{{k_2}-{k_1}}$$
Let us check the result using $k_1=\log(2015)$ and $k_2=\log(2014)$ and let us compute the value of the integral between $0$ and $100$. The values obtained exactly match the results of the numerical eveluation of the original integral (for six significant figures).
If this approximation is correct, it could easily be extended to the calculation of $$I_n=\int \frac{dx}{\sum _{i=1}^n e^{k_i x }} \simeq \int\frac{e^{-k_1x}}{n+Ax }dx=\frac{e^{\frac{{k_1} n}{A}} \text{Ei}\left(-\frac{{k_1} (n+A
   x)}{A}\right)}{A}$$ where $A=\sum_{i=1}^n (k_i -k_1)$
