A calculus problem by Jose Sacramento

Relevant wiki: Integration Tricks

$\begin{aligned} I & = \int_{-1}^1 \frac 1 {1+|x|^x} dx & \small \color{#3D99F6} \text{Using the identity: }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_{-1}^1 \frac 1{1+|x|^x} + \frac 1{1+|-x|^{-x}} dx & \small \color{#3D99F6} \text{Note that } |-x| = |x| \\ & = \frac 12 \int_{-1}^1 \frac 1{1+|x|^x} + \frac {|x|^x}{|x|^x+1} dx \\ & = \frac 12 \int_{-1}^1 1 \cdot dx = \frac 12 x \ \bigg|_{-1}^1 = \boxed{1} \end{aligned}$

Since $|x|=|-x|$ we have $\displaystyle I=\int_{-1}^{1} \dfrac{dx} {1+|x|^x}=\int_{-1}^{1} \dfrac{dx} {1+|x|^{-x}}$

Adding them we have $\displaystyle 2I=\int_{-1}^{1} dx =2$ . Thus the answer is $\boxed{1}$