1. **State the problem:** We want to find the integral $$\int \ln\left(x + \sqrt{x^2 - 9}\right) \, dx.$$\n\n2. **Recall the formula and approach:** Integration by parts is useful for integrals involving logarithms. The formula is $$\int u \, dv = uv - \int v \, du.$$\n\n3. **Choose parts:** Let $$u = \ln\left(x + \sqrt{x^2 - 9}\right)$$ and $$dv = dx.$$ Then, $$du = \frac{d}{dx} \ln\left(x + \sqrt{x^2 - 9}\right) dx$$ and $$v = x.$$\n\n4. **Compute $$du$$:** Using the chain rule,\n$$du = \frac{1}{x + \sqrt{x^2 - 9}} \left(1 + \frac{x}{\sqrt{x^2 - 9}}\right) dx = \frac{1 + \frac{x}{\sqrt{x^2 - 9}}}{x + \sqrt{x^2 - 9}} dx.$$\nSimplify numerator:\n$$1 + \frac{x}{\sqrt{x^2 - 9}} = \frac{\sqrt{x^2 - 9} + x}{\sqrt{x^2 - 9}}.$$\nSo,\n$$du = \frac{\frac{\sqrt{x^2 - 9} + x}{\sqrt{x^2 - 9}}}{x + \sqrt{x^2 - 9}} dx = \frac{\sqrt{x^2 - 9} + x}{\sqrt{x^2 - 9} (x + \sqrt{x^2 - 9})} dx.$$\n\n5. **Simplify $$du$$:** Notice numerator and denominator have $$x + \sqrt{x^2 - 9}$$ which are the same, so\n$$du = \frac{\cancel{\sqrt{x^2 - 9} + x}}{\sqrt{x^2 - 9} \cancel{(x + \sqrt{x^2 - 9})}} dx = \frac{1}{\sqrt{x^2 - 9}} dx.$$\n\n6. **Apply integration by parts:**\n$$\int \ln\left(x + \sqrt{x^2 - 9}\right) dx = x \ln\left(x + \sqrt{x^2 - 9}\right) - \int x \cdot \frac{1}{\sqrt{x^2 - 9}} dx.$$\n\n7. **Evaluate the remaining integral:**\n$$\int \frac{x}{\sqrt{x^2 - 9}} dx.$$\nUse substitution $$t = x^2 - 9,$$ so $$dt = 2x dx,$$ hence $$x dx = \frac{dt}{2}.$$\n\n8. **Rewrite integral:**\n$$\int \frac{x}{\sqrt{x^2 - 9}} dx = \int \frac{1}{\sqrt{t}} \cdot \frac{dt}{2} = \frac{1}{2} \int t^{-1/2} dt = \frac{1}{2} \cdot 2 t^{1/2} + C = \sqrt{t} + C = \sqrt{x^2 - 9} + C.$$\n\n9. **Final answer:**\n$$\int \ln\left(x + \sqrt{x^2 - 9}\right) dx = x \ln\left(x + \sqrt{x^2 - 9}\right) - \sqrt{x^2 - 9} + C.$$