Answer
The statement is false:
$$
\int_{0}^{4} \frac{x}{x^{2}-1} d x
$$
Observe that the given integral is improper because $ f(x)= \frac{x}{x^{2}-1}$ has the vertical asymptote $x=1$. The infinite discontinuity occurs at the middle of the interval $ [0,4] $ at $x = 1$,
The given improper integral
$$
\int_{0}^{4} \frac{x}{x^{2}-1} d x= \infty
$$
is divergent.
Work Step by Step
$$
\int_{0}^{4} \frac{x}{x^{2}-1} d x
$$
Observe that the given integral is improper because $ f(x)= \frac{x}{x^{2}-1}$ has the vertical asymptote $x=1$. Since the infinite discontinuity occurs at the middle of the interval $ [0,4] $ at $x = 1$, we must use part (c) of Definition 3 with $ c=1 $:
$$
\int_{0}^{4} \frac{x}{x^{2}-1} d x=\int_{0}^{1} \frac{x}{x^{2}-1} d x +\int_{1}^{4} \frac{x}{x^{2}-1} d x
$$
where
$$
\begin{split}
\int_{0}^{1} \frac{x}{x^{2}-1} d x& = \lim _{t \rightarrow 1^{-}} \int_{0}^{t} \frac{x}{x^{2}-1} d x
\\
& =\lim _{t \rightarrow 1^{-}}\left[\frac{1}{2} \ln \left|x^{2}-1\right|\right]_{0}^{t}
\\
& =\lim _{t \rightarrow 1^{-}} \frac{1}{2} \ln \left|t^{2}-1\right| \\
& =\infty,
\end{split}
$$
so the integral
$$
\int_{0}^{1} \frac{x}{x^{2}-1} d x
$$
is divergent, and the given improper integral,
$$
\int_{0}^{4} \frac{x}{x^{2}-1} d x
$$
is divergent.
Thus the statement is False.
