To solve \(\int_{1}^{\infty} \frac{1}{(3x - 2)^{\frac{3}{2}}} \, dx\), we first perform a substitution. Let \(u = 3x - 2\), then \(du = 3 \, dx\) or \(dx = \frac{du}{3}\).
The limits of integration change as follows: when \(x = 1\), \(u = 1\), and as \(x \to \infty\), \(u \to \infty\).
The integral becomes:
\(\int_{1}^{\infty} \frac{1}{u^{\frac{3}{2}}} \cdot \frac{1}{3} \, du = \frac{1}{3} \int_{1}^{\infty} u^{-\frac{3}{2}} \, du\)
Integrating \(u^{-\frac{3}{2}}\), we get:
\(\frac{1}{3} \left[ \frac{u^{-\frac{1}{2}}}{-\frac{1}{2}} \right]_{1}^{\infty} = \frac{1}{3} \left[ -2u^{-\frac{1}{2}} \right]_{1}^{\infty}\)
Evaluating the limits:
\(\frac{1}{3} \left[ -2(\infty)^{-\frac{1}{2}} + 2(1)^{-\frac{1}{2}} \right] = \frac{1}{3} \left[ 0 + 2 \right] = \frac{2}{3}\)
