# Improper Integral

• Oct 21st 2009, 05:47 PM
WhoCares357
Improper Integral
$\displaystyle \int^3_{-\infty}\frac{dx}{x^2+9}$
I got to the part where $\displaystyle \lim_{b->-\infty}\frac{1}{3}tan^{-1}(1)-\frac{1}{3}tan^{-1}(\frac{b}{3})$
I'm not sure how to evaluate arctan to -infinity...

- Wolfram|Alpha[1%2F(x^2%2B9)%2C[x]]
- Wolfram|Alpha[1%2F%28x^2%2B9%29%2C[x%2C-infinity%2C3]]
• Oct 21st 2009, 05:50 PM
skeeter
Quote:

Originally Posted by WhoCares357
$\displaystyle \int ^3_{-\infty}\frac{dx}{x^2+9)$
I got to the part where $\displaystyle \lim_{b->-\infty}\frac{1}{3}tan^{-1}(1)-\frac{1}{3}tan^{-1}(\frac{b}{3})$
I'm not sure how to evaluate arctan to -infinity...

- Wolfram|Alpha[1%2F(x^2%2B9)%2C[x]]
- Wolfram|Alpha[1%2F%28x^2%2B9%29%2C[x%2C-infinity%2C3]]

familiar with the graph of the arctangent function?

$\displaystyle \lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}$
• Oct 21st 2009, 05:53 PM
WhoCares357
Quote:

Originally Posted by skeeter
familiar with the graph of the arctangent function?

$\displaystyle \lim_{x \to -\infty} \arctan(x) = -\frac{\pi}{2}$

Thanks.