Limits with Integration

To be precise,

The expression $\dfrac{e^t -1}{e^t+1}$ is an odd function.

We know that integration of odd function with limits of the kind $-x \space to \space x$ is $\boxed{0}$

The above integrand can be expressed as:

$\frac{e^t - 1}{e^t + 1} = \frac{\frac{e^t - 1}{2e^{t/2}}}{\frac{e^t+1}{2e^{t/2}}} = \frac{\frac{e^{t/2} - e^{-t/2}}{2}}{\frac{e^{t/2} + e^{-t/2}}{2}} = \frac{sinh(\frac{t}{2})}{cosh(\frac{t}{2})} = tanh(\frac{t}{2}).$

Integrating this with respect to t yields $2 ln(cosh(\frac{t}{2}))$ , and if we supply the limits $x$ and $-x$ we now get:

$2 \cdot [ln(cosh(\frac{x}{2}) - ln(cosh(\frac{-x}{2})] = 0$ for all $x \in \mathbb{R}.$

The value of the problem must be zero.