Solution
= Solution
The original PDF places the $-1$ outside the square root. Use the <substitution rule> with $t=\sqrt x$, so $dx=2t\,dt$ and the endpoints become $2$ and $3$. Polynomial division gives
$$
\int_2^3\frac{2t}{t-1}\,dt=2\int_2^3\left(1+\frac1{t-1}\right)dt=2[t+\ln(t-1)]_2^3.
$$
Hence the <definite integral> is
$$
\boxed{2+2\ln2.}
$$