1. **State the problem:** We need to evaluate the integral $$\int (x + 2) \sqrt{x^2 + 10x - 11} \, dx.$$\n\n2. **Identify the integral type and substitution:** The integrand involves a polynomial times the square root of a quadratic expression. A common method is to use substitution for the quadratic inside the square root. Let $$u = x^2 + 10x - 11.$$\n\n3. **Compute the derivative of $u$:**\n$$\frac{du}{dx} = 2x + 10.$$\n\n4. **Rewrite the integrand in terms of $u$ and $du$:**\nWe have $x + 2$ in the integrand, but $du/dx = 2x + 10$. Notice that $x + 2$ is related to $\frac{du}{dx}$ by\n$$x + 2 = \frac{1}{2}(2x + 4) = \frac{1}{2}(du/dx - 6).$$\n\n5. **Express the integral in terms of $u$ and $du$:**\n$$\int (x + 2) \sqrt{u} \, dx = \int \frac{1}{2}(du/dx - 6) \sqrt{u} \, dx = \frac{1}{2} \int (du - 6 dx) \sqrt{u}.$$\n\n6. **Express $dx$ in terms of $du$ and $x$:**\nFrom $du = (2x + 10) dx$, we get\n$$dx = \frac{du}{2x + 10}.$$\n\n7. **Rewrite the integral carefully:**\n$$\int (x + 2) \sqrt{u} \, dx = \int (x + 2) \sqrt{u} \, dx = \int \frac{x + 2}{2x + 10} \sqrt{u} \, du.$$\n\n8. **Simplify the fraction:**\n$$\frac{x + 2}{2x + 10} = \frac{x + 2}{2(x + 5)} = \frac{1}{2} \cdot \frac{x + 2}{x + 5}.$$\n\n9. **Rewrite the integral:**\n$$\int (x + 2) \sqrt{u} \, dx = \frac{1}{2} \int \frac{x + 2}{x + 5} \sqrt{u} \, du.$$\n\n10. **Express $x$ in terms of $u$ to eliminate $x$:**\nRecall $u = x^2 + 10x - 11 = (x + 5)^2 - 36$, so\n$$x + 5 = \sqrt{u + 36}.$$\nThen\n$$x + 2 = (x + 5) - 3 = \sqrt{u + 36} - 3.$$\n\n11. **Substitute back:**\n$$\frac{x + 2}{x + 5} = \frac{\sqrt{u + 36} - 3}{\sqrt{u + 36}} = 1 - \frac{3}{\sqrt{u + 36}}.$$\n\n12. **Rewrite the integral in $u$ only:**\n$$\frac{1}{2} \int \left(1 - \frac{3}{\sqrt{u + 36}}\right) \sqrt{u} \, du = \frac{1}{2} \int \sqrt{u} \, du - \frac{3}{2} \int \frac{\sqrt{u}}{\sqrt{u + 36}} \, du.$$\n\n13. **Evaluate the first integral:**\n$$\int \sqrt{u} \, du = \int u^{1/2} \, du = \frac{2}{3} u^{3/2} + C.$$\n\n14. **Evaluate the second integral:**\nRewrite the integrand:\n$$\frac{\sqrt{u}}{\sqrt{u + 36}} = \sqrt{\frac{u}{u + 36}}.$$\nLet $$w = \sqrt{\frac{u}{u + 36}}.$$\nThis integral is more complicated; instead, use substitution $u = 36 \tan^2 \theta$:\nThen $u + 36 = 36 \sec^2 \theta$, so\n$$\sqrt{\frac{u}{u + 36}} = \sqrt{\frac{36 \tan^2 \theta}{36 \sec^2 \theta}} = \sqrt{\tan^2 \theta \cos^2 \theta} = \tan \theta \cos \theta = \sin \theta.$$\nAlso,\n$$du = 72 \tan \theta \sec^2 \theta \, d\theta.$$\n\n15. **Rewrite the integral:**\n$$\int \sqrt{\frac{u}{u + 36}} \, du = \int \sin \theta \cdot 72 \tan \theta \sec^2 \theta \, d\theta = 72 \int \sin \theta \tan \theta \sec^2 \theta \, d\theta.$$\n\n16. **Simplify the integrand:**\nRecall $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\sec \theta = \frac{1}{\cos \theta}$, so\n$$\sin \theta \tan \theta \sec^2 \theta = \sin \theta \cdot \frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\cos^2 \theta} = \frac{\sin^2 \theta}{\cos^3 \theta}.$$\n\n17. **Integral becomes:**\n$$72 \int \frac{\sin^2 \theta}{\cos^3 \theta} \, d\theta = 72 \int \frac{1 - \cos^2 \theta}{\cos^3 \theta} \, d\theta = 72 \int \left(\frac{1}{\cos^3 \theta} - \frac{\cos^2 \theta}{\cos^3 \theta}\right) d\theta = 72 \int (\sec^3 \theta - \sec \theta) \, d\theta.$$\n\n18. **Use known integrals:**\n$$\int \sec \theta \, d\theta = \ln |\sec \theta + \tan \theta| + C,$$\n$$\int \sec^3 \theta \, d\theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln |\sec \theta + \tan \theta| + C.$$\n\n19. **Evaluate:**\n$$72 \int (\sec^3 \theta - \sec \theta) \, d\theta = 72 \left(\frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln |\sec \theta + \tan \theta| - \ln |\sec \theta + \tan \theta|\right) + C = 36 \sec \theta \tan \theta - 36 \ln |\sec \theta + \tan \theta| + C.$$\n\n20. **Back-substitute $\theta$ to $u$:**\nRecall $u = 36 \tan^2 \theta$, so\n$$\tan \theta = \frac{\sqrt{u}}{6}, \quad \sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + \frac{u}{36}} = \frac{\sqrt{u + 36}}{6}.$$\n\n21. **Substitute back:**\n$$36 \sec \theta \tan \theta = 36 \cdot \frac{\sqrt{u + 36}}{6} \cdot \frac{\sqrt{u}}{6} = \sqrt{u} \sqrt{u + 36}.$$\n$$36 \ln |\sec \theta + \tan \theta| = 36 \ln \left| \frac{\sqrt{u + 36} + \sqrt{u}}{6} \right| = 36 \ln |\sqrt{u + 36} + \sqrt{u}| - 36 \ln 6.$$\n\n22. **Combine all parts:**\nThe original integral is\n$$\frac{1}{2} \cdot \frac{2}{3} u^{3/2} - \frac{3}{2} \left( \sqrt{u} \sqrt{u + 36} - 36 \ln |\sqrt{u + 36} + \sqrt{u}| \right) + C = \frac{1}{3} u^{3/2} - \frac{3}{2} \sqrt{u} \sqrt{u + 36} + 54 \ln |\sqrt{u + 36} + \sqrt{u}| + C'.$$\n\n23. **Replace $u$ by $x^2 + 10x - 11$:**\n$$\boxed{\int (x + 2) \sqrt{x^2 + 10x - 11} \, dx = \frac{1}{3} (x^2 + 10x - 11)^{3/2} - \frac{3}{2} \sqrt{x^2 + 10x - 11} \sqrt{x^2 + 10x + 25} + 54 \ln \left| \sqrt{x^2 + 10x + 25} + \sqrt{x^2 + 10x - 11} \right| + C}.$$