Answer
$$
\int_{0}^{5} \frac{w}{w-2} d w
$$
We note first that the given integral is improper because $ f(x)= \frac{w}{w-2}$ has the vertical asymptote $w=2 $. The infinite discontinuity occurs at the middle of the interval $ [0,5] $ at $w = 2$.
The given improper integral is divergent.
Work Step by Step
$$
\int_{0}^{5} \frac{w}{w-2} d w
$$
Observe that the given integral is improper because $ f(x)= \frac{w}{w-2}$ has the vertical asymptote $w=2 $. Since the infinite discontinuity occurs at the middle of the interval $ [0,5] $ at $w = 2$, we must use part (c) of Definition 3 with $ c=2 $:
$$
\int_{0}^{5} \frac{w}{w-2} d w=\int_{0}^{2} \frac{w}{w-2} d w+\int_{2}^{5} \frac{w}{w-2} d w
$$
where
$$
\begin{split}
\int_{0}^{2} \frac{w}{w-2} d w & = \lim _{t \rightarrow 2^{-}} \int_{0}^{t} \frac{w}{w-2} d w \\
& =\lim _{t \rightarrow 2^{-}} \int_{0}^{t} \frac{w-2+2}{w-2} d w \\
& =\lim _{t \rightarrow 2^{-}} \int_{0}^{t}\left(1+\frac{2}{w-2}\right) d w \\
& =\lim _{t \rightarrow 2^{-}}[w+2 \ln |w-2|]_{0}^{t} \\
& =\lim _{t \rightarrow 2^{-}}(t+2 \ln |t-2|-2 \ln 2) \\
& =-\infty,
\end{split}
$$
so the integral
$$
\int_{0}^{2} \frac{w}{w-2} d w
$$
is divergent , and hence,
$$
\int_{0}^{5} \frac{w}{w-2} d w
$$
is divergent .
Thus the given improper integral is divergent.