Answer: \(\ln\frac53-\frac{52}{225}\).
We start with the main method. Simplify the integrand before integrating, then substitute the upper and lower limits carefully.
First expand the integrand:
\(\frac{(x-1)^2}{x^3}=\frac{x^2-2x+1}{x^3}=\frac1x-\frac2{x^2}+\frac1{x^3}.\)
So
\(\int_3^5 \frac{(x-1)^2}{x^3}\,\mathrm dx =\int_3^5\left(x^{-1}-2x^{-2}+x^{-3}\right)\mathrm dx.\)
Integrating term by term gives
\(\left[\ln x+\frac2x-\frac1{2x^2}\right]_3^5.\)
At \(x=5\), the value is
\(\ln5+\frac25-\frac1{50}.\)
At \(x=3\), the value is
\(\ln3+\frac23-\frac1{18}.\)
Therefore the exact value is
\(\left(\ln5+\frac25-\frac1{50}\right)-\left(\ln3+\frac23-\frac1{18}\right).\)
The logarithmic part is
\(\ln5-\ln3=\ln\frac53.\)
The numerical part is
\(\frac25-\frac1{50}-\frac23+\frac1{18}=-\frac{52}{225}.\)
Hence
\(\displaystyle \int_3^5 \frac{(x-1)^2}{x^3}\,\mathrm dx=\ln\frac53-\frac{52}{225}.\)
This completes the solution and gives the required result.
