So I have a two fold question, one I believe is simple but my algebra seems to be off, the other involves the trapezoidal rule of integration using Mathematica as an aid. Here they are:
$1.\quad \displaystyle \int_{-\infty}^{\infty} \frac{\operatorname{sech}(x)}{x^2+1} dx = \int_{-1}^{1} \operatorname{sech}\left(\frac{1}{1-t^2}\right)\frac{t^2+1}{t^4-t^2+1} dt$
I know I need to let $x = \frac{t}{1-t^2}$ and take the limits as $t \to \infty$,change the limits of integration and do the same for $t \to -\infty$ but I can't seem to nail it down. Why are my limits going to be $-1$ and $1$?
$2$. Space five points equally from $-1$ to $1$ and compute the four trapezoid approximation of $\int_{-1}^{1} \mathrm{sech}(\frac{1}{1-t^2})\frac{t^2+1}{t^4-t^2+1} dt$ using Mathematica to evaluate $\operatorname{sech}(x)$. To be honest, I'm not really sure what the question is asking. Am I breaking the integral up into four integrations the first of which is from $-1$ to $-0.5$? How do I use Mathematica to evaluate? Any help is appreciated.
$\quad \displaystyle \int_{-1}^{1} \mathrm{sech}\left(\frac{1}{1-t^2}\right)\frac{t^2+1}{t^4-t^2+1} dt = \lim n \to \infty \frac{b-a}{2n}[f(x_0) + 2f(x_1) + 2f(x_2) + \cdots + 2f(x_{(n-1)}) + f(x_n)] $