Question

Use limits of definite integrals to calculate each of the improper integrals in Exercises 21–56.

${\int }_0^{\infty }\frac{x}{\left(x^2+1\right)}dx$

Short answer

The value is$\infty$.

Step-by-step solution

Step 1. Given information.

The given function is${\int }_0^{\infty }\frac{x}{\left(x^2+1\right)}dx$.

Step 2. Value of the integral.

The given integral can be written as,

${\int }_0^{\infty }\frac{x}{\left(x^2+1\right)}dx=\underset{B\to \infty }{\lim }{\int }_0^B\frac{x}{\left(x^2+1\right)}dx$

$$\begin{gathered} \text{Let}u=x^2+1\text{i.e}du=2xdx \\ Therefore, \\ \int \frac{x}{\left(x^2+1\right)}dx=\int \frac{1}{2u}du \\ =\frac{1}{2}\int \frac{1}{u}du \\ =\frac{1}{2}\ln \left|u\right| \\ \int \frac{x}{\left(x^2+1\right)}dx=\frac{1}{2}\ln \left|x^2+1\right| \end{gathered}$$

Step 3. Substitution.

Substituting the obtained value in the given equation, we get,

$$\begin{gathered} \int \frac{x}{\left(x^2+1\right)}dx=\frac{1}{2}\ln \left|x^2+1\right| \\ {\int }_0^{\infty }\frac{x}{\left(x^2+1\right)}dx=\underset{B\to \infty }{\lim }\frac{1}{2}\ln \left|x^2+1\right| \\ =\infty -0 \\ =\infty \end{gathered}$$