Chapter 9.4, Problem 14E

To determine

To find: the value of improper integral.

Answer to Problem 14E

The value of integral is $\frac{1}{2}\text{ln3}$ .

Explanation of Solution

Given information:

The improper integral is ${\displaystyle {\int }_{-\infty }^0\frac{1}{x^2-4x+3}dx}$ .

Calculation: by using definition of improper integrals with infinite integration limits, the improper integral can be written as:

  ${\displaystyle {\int }_{-\infty }^0\frac{1}{x^2-4x+3}dx}=\underset{a\to -\infty }{\lim }{\displaystyle {\int }_a^0\frac{1}{x^2-4x+3}dx}$

To find the value of integral, first integrate and then simplify,

Thus, the value is $\frac{1}{2}\text{ln3}$ .