Answer
$$
\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x
$$
We note first that the given integral is improper because $ f(x)= \frac{1}{\sqrt[3]{x-1}}$ has the vertical asymptote $x=1 $. The infinite discontinuity occurs at the middle of the interval $ [0,9] $ at $x = 1$.
The given improper integral is convergent and its value is
$$
\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x = \frac{9}{2}.
$$
Work Step by Step
$$
\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x
$$
Observe that the given integral is improper because $ f(x)= \frac{1}{\sqrt[3]{x-1}}$ has the vertical asymptote $x=1 $. Since the infinite discontinuity occurs at the middle of the interval $ [0,9] $ at $x = 1$, we must use part (c) of Definition 3 with $ c=1$:
$$
\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x=\int_{0}^{1}(x-1)^{-1 / 3} d x+\int_{1}^{9}(x-1)^{-1 / 3} d x \quad (1)
$$
where
$$
\begin{split}
\int_{0}^{1}(x-1)^{-1 / 3} d x & = \lim _{t \rightarrow 1^{-}} \int_{0}^{t}(x-1)^{-1 / 3} d x \\
& =\lim _{t \rightarrow 1^{-}}\left[\frac{3}{2}(x-1)^{2 / 3}\right]_{0}^{t} \\
& =\lim _{t \rightarrow 1^{-}}\left[\frac{3}{2}(t-1)^{2 / 3}-\frac{3}{2}\right] \\
& =-\frac{3}{2}
\end{split} \quad \quad\quad (2)
$$
and
$$
\begin{split}
\int_{1}^{9}(x-1)^{-1 / 3} d x & = \lim _{t \rightarrow 1^{+}} \int_{t}^{9}(x-1)^{-1 / 3} d x \\
& =\lim _{t \rightarrow 1^{+}}\left[\frac{3}{2}(x-1)^{2 / 3}\right]_{t}^{9} \\
& =\lim _{t \rightarrow 1^{+}}\left[6-\frac{3}{2}(t-1)^{2 / 3}\right] \\
& =6
\end{split} \quad \quad\quad (3)
$$
Substituting the integrals (2) and (3) into expression (1) , we obtain
$$
\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x=-\frac{3}{2} +6 =\frac{9}{2},
$$
It follows that the integral
$$
\int_{0}^{9} \frac{1}{\sqrt[3]{x-1}} d x
$$
is convergent.
