Chapter 5.10, Problem 15E

To determine

To find whether the integral ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{x+1}{x^2+2x}dx}$ is convergent or divergent.

Answer to Problem 15E

The integral ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{x+1}{x^2+2x}dx}$ is convergent.

Explanation of Solution

Given information:

The integral is ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{x+1}{x^2+2x}dx}$

Concept used: If the value of the integral is finite, it is convergent otherwise it is divergent

Given, ${\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{x+1}{x^2+2x}dx}$

On solving the function of the integral:

  $={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{x+1}{x(x+2)}dx=}{\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{x+2-1}{x(x+2)}dx}$

By factorization, given integral becomes as:

  $\begin{array}{l}={\displaystyle \underset{1}{\overset{\infty }{\int }}\left(\frac{x+2}{x(x+2)}-\frac{1}{x(x+2)}\right)dx}\\ ={\displaystyle \underset{1}{\overset{\infty }{\int }}\left(\frac{1}{x}-\frac{1}{2}\left(\frac{1}{x}-\frac{1}{(x+2)}\right)\right)dx}\\ ={\displaystyle \underset{1}{\overset{\infty }{\int }}\frac{1}{2}\left(\frac{1}{x}-\frac{1}{(x+2)}\right)dx}\end{array}$

On Integration, the integral becomes

  $\begin{array}{l}\text{=}\frac{1}{2}{\left[\ln (x)-\ln (x+2)\right]}_1^{\infty }=\frac{1}{2}{\left[\ln \left(\frac{x}{x+2}\right)\right]}_1^{\infty }\\ =\frac{1}{2}{\left[\ln \left(\frac{x}{x(1+2/x)}\right)\right]}_1^{\infty }=\frac{1}{2}\left[0-\ln \left(\frac{1}{1+2}\right)\right]\text{[cancel }x\text{ from numerator and denominator]}\\ =\frac{1}{2}\left[\ln \left(3\right)\right],\text{a}\text{finite}\text{number}\end{array}$

Since the given integral has finite value.

Hence, the given integral is convergent