1. ## Improper Integrals

I can't figure out this one...

Let f(x) = $\frac{x^2 + 2}{x^2 + 1}$

a) Show that f(x) > 1 for all x

b) Use the inequality in part (a) to show that $\int_{-\infty}^{\infty}f(x)dx$ diverges.

2. Originally Posted by larson
I can't figure out this one...

Let f(x) = $\frac{x^2 + 2}{x^2 + 1}$

a) Show that f(x) > 1 for all x

b) Use the inequality in part (a) to show that $\int_{-\infty}^{\infty}f(x)dx$ diverges.

(a) $\frac{x^2 + 2}{x^2 + 1} = \frac{(x^2 + 1) + 1}{x^2 + 1} = 1 + \frac{1}{x^2 + 1}$ .....
(b) $\int_{-\infty}^{\infty}f(x) \, dx > \int_{-\infty}^{\infty}1 \, dx$ .......
(a) $\frac{x^2 + 2}{x^2 + 1} = \frac{(x^2 + 1) + 1}{x^2 + 1} = 1 + \frac{1}{x^2 + 1}$ .....
(b) $\int_{-\infty}^{\infty}f(x) \, dx > \int_{-\infty}^{\infty}1 \, dx$ .......
as Mr. Fantastic showed you you can rewrite his as $\int_{-\infty}^{\infty}1+\frac{1}{x^2+1}dx$...which is $\bigg[x+arctan(x)\bigg]\bigg|_{-\infty}^{\infty}=\bigg[{\infty}+\frac{\pi}{2}\bigg]-\bigg[-{\infty}-\frac{\pi}{2}\bigg]\Rightarrow{divergent}$