Science:Math Exam Resources/Courses/MATH101/April 2012/Question 03 (d)/Solution 1

The integral is equal to the sum of the two standard Type I integrals:

$\underset{-\mathrm{\infty }}{\overset{0}{\int }}\frac{x}{x^2+1}dx+\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{x^2+1}dx$.

By the substitution u=x²+1, du = 2dx, we see that

$\begin{array}{rl}\int \frac{x}{x^2+1}dx& =\frac{1}{2}\int \frac{1}{u}du\\ & =\frac{1}{2}\ln |u|+C\\ & =\frac{1}{2}\ln |x^2+1|+C\end{array}$

So, we calculate

$\begin{array}{rl}\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{x^2+1}dx& =\underset{a\to \mathrm{\infty }}{\lim }\frac{1}{2}\ln (x^2+1){|}_0^a\\ & =\underset{a\to \mathrm{\infty }}{\lim }(\frac{1}{2}\ln (a^2+1)-\ln (1))\\ & =\mathrm{\infty }\end{array}$

Since this part diverges the whole integral diverges also.