How to calculate the integral $\int_{0}^{1}e^{\arctan x}\,dx$? How to calculate the integral
$$I=\int_{0}^{1}e^{\arctan x}\,dx\quad?$$
My attempts using the partitioning method or the variable change method did not result in the desired result.
Thank you!
 A: With CAS help and with substitution: $x=\tan (t)$ we have:
$$\int_{0}^{1}e^{\arctan x}\,dx=\\\int_0^{\frac{\pi }{4}} \frac{\exp (t)}{\cos ^2(t)} \, dt=\\\left(\frac{8}{5}+\frac{4
   i}{5}\right) e^{\pi /4} \, _2F_1\left(1-\frac{i}{2},2;2-\frac{i}{2};-i\right)+\frac{1}{2} \left(2 i+\psi
   \left(\frac{1}{2}-\frac{i}{4}\right)-\psi \left(1-\frac{i}{4}\right)\right)\\\approx 1.5913$$
where:$\, _2F_1\left(1-\frac{i}{2},2;2-\frac{i}{2};-i\right)$ is hypergeometric function,$\psi \left(\frac{1}{2}-\frac{i}{4}\right)$ is digamma function.
A: As Tolaso commented, no hope to have a closed form solution in terms of elementary functions.
If you did plot the integrand over the given range, you must have noticed that it is very close to a straight line and, more than likely, a series solution would be sufficient.
If you use  a series solution, you should get
$$e^{\tan ^{-1}(x)}=\sum_{n=0}^\infty a_n x^n$$ where the $a_n$ are defined by
$$a_n=\frac{a_{n-1}-(n-2)\, a_{n-2}}{n}\qquad \text{with}\qquad a_0=a_1=1$$ you should end with
$$\int_0^1 e^{\tan ^{-1}(x)}\,dx=\sum_{n=0}^\infty \frac {a_n }{n+1}$$ Just use the first terms to get more than reasonable approximations of the result.
