Answer
False
Work Step by Step
Given: $ \int^{4}_{0}\frac{x}{x^{2}-1}dx$
Since, the $\frac{x}{x^{2}-1}dx$ is bounded at $x=1$
Therefore,
$ \int^{4}_{0}\frac{x}{x^{2}-1}dx=\int^{1}_{0}\frac{x}{x^{2}-1}d+ \int^{4}_{1}\frac{x}{x^{2}-1}dx$
$=\frac{1}{2}\int^{1}_{0}\frac{2x}{x^{2}-1}dx+ \frac{1}{2}\int^{4}_{1}\frac{2x}{x^{2}-1}dx$
$=[\frac{1}{2}ln(x^{2}-1]^{1}_{0}+[\frac{1}{2}ln(x^{2}-1]^{4}_{1}$
$=-\infty+\infty$
= Indeterminate
Hence, the given integral diverges and is not equal to $\frac{1}{2}ln15$.
