### Home > CALC > Chapter 10 > Lesson 10.4.1 > Problem10-138

10-138.

$=\frac{1}{25}\int_{-\infty}^{\infty}\frac{1}{(x/5)^2+1}dx$

$=\lim_{a\to-\infty}\frac{1}{25}\int_{a}^{0}\frac{1}{(x/5)^2+1}dx+\lim_{b\to\infty}\frac{1}{25}\int_{0}^{b}\frac{1}{(x/5)^2+1}dx$

Use substitution. Let u = 16 – x^2.

A limit is needed to evaluate this integral.

$=\lim_{c\to-\infty}\int_{c}^{-1}x^{-3}dx$

$\frac{1}{x^2-2x-3}=\frac{-1/4}{x+1}+\frac{1/4}{x-3}$