Evaluate the integrals in Exercises 31–78.
58. ∫(from 0 to ln9)e^θ(e^θ-1)^(1/2) dθ

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Start by examining the integral: \(\int_0^{\ln 9} e^{\theta} \left(e^{\theta} - 1\right)^{1/2} \, d\theta\). Notice that the integrand contains \(e^{\theta}\) and a function of \(e^{\theta} - 1\), which suggests a substitution involving \(e^{\theta}\).

Let’s perform the substitution \(u = e^{\theta} - 1\). Then, differentiate both sides with respect to \(\theta\) to find \(du\): \(du = e^{\theta} d\theta\).

Rewrite the integral in terms of \(u\). Since \(du = e^{\theta} d\theta\), the integral becomes \(\int \sqrt{u} \, du\) after changing the limits accordingly.

Change the limits of integration from \(\theta\) to \(u\): when \(\theta = 0\), \(u = e^0 - 1 = 0\); when \(\theta = \ln 9\), \(u = e^{\ln 9} - 1 = 9 - 1 = 8\).

Now, the integral is \(\int_0^8 u^{1/2} \, du\). This is a standard power integral, which can be integrated using the formula \(\int u^n du = \frac{u^{n+1}}{n+1} + C\) for \(n \neq -1\).

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