To solve \(\int_{1}^{8} x^{-\frac{2}{3}} \ln x \, dx\), we use integration by parts.
Let \(u = \ln x\) and \(dv = x^{-\frac{2}{3}} \, dx\).
Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^{\frac{1}{3}}}{\frac{1}{3}} = 3x^{\frac{1}{3}}\).
Integration by parts gives:
\(\int u \, dv = uv - \int v \, du\)
\(= 3x^{\frac{1}{3}} \ln x - 3 \int x^{\frac{1}{3}} \cdot \frac{1}{x} \, dx\)
\(= 3x^{\frac{1}{3}} \ln x - 3 \int x^{-\frac{2}{3}} \, dx\)
Integrating \(x^{-\frac{2}{3}}\), we get:
\(\int x^{-\frac{2}{3}} \, dx = \frac{x^{\frac{1}{3}}}{\frac{1}{3}} = 3x^{\frac{1}{3}}\)
Thus, the integral becomes:
\(3x^{\frac{1}{3}} \ln x - 9x^{\frac{1}{3}}\)
Evaluate from 1 to 8:
\(\left[ 3x^{\frac{1}{3}} \ln x - 9x^{\frac{1}{3}} \right]_{1}^{8}\)
\(= \left( 3 \cdot 8^{\frac{1}{3}} \ln 8 - 9 \cdot 8^{\frac{1}{3}} \right) - \left( 3 \cdot 1^{\frac{1}{3}} \ln 1 - 9 \cdot 1^{\frac{1}{3}} \right)\)
\(= \left( 6 \ln 8 - 18 \right) - \left( 0 - 9 \right)\)
\(= 6 \ln 8 - 18 + 9\)
\(= 6 \ln 8 - 9\)
Since \(\ln 8 = 3 \ln 2\), we have:
\(6 \ln 8 = 18 \ln 2\)
Thus, the final result is:
\(18 \ln 2 - 9\)
