Improper Integral

Although the denominator may look simplified, in reality there is a negative exponent in it. Our first step should be to remove that:

$\int_0^\infty \frac{1}{e^x + e^{-x}} \, dx = \int_0^\infty \frac{1}{e^x + \tfrac{1}{e^x}} \, dx = \int_0^\infty \frac{e^x}{e^{2x} + 1} \, dx.$

Now, it looks like we can group the numerator with the differential $dx$ and use a clever $u$ -substitution to simplify the problem. Choosing $u = e^x$ (and $du = e^x \, dx$ ), the problem continues:

$\int_0^\infty \frac{e^x}{e^{2x} + 1} \, dx = \int_0^\infty \frac{1}{e^{2x} + 1} \cdot e^x \, dx = \int_1^\infty \frac{1}{u^2 + 1} \, du.$

Now, the trigonometric substitution $u = \tan\theta$ (and $du = \sec^2\theta \, d\theta$ ) allows us to finish the problem:

$\int_1^\infty \frac{1}{u^2 + 1} \, du = \int_{\pi/4}^{\pi/2} \frac{1}{\tan^2\theta + 1} \cdot \sec^2\theta \, d\theta = \int_{\pi/4}^{\pi/2} \frac{1}{sec^2\theta} \cdot \sec^2\theta \, d\theta = \int_{\pi/4}^{\pi/2} 1 \, d\theta = \boxed{\frac{\pi}{4}.}$