integration 25

A simpler solution suggested by @Pi Han Goh using hyperbolic trigonometric function substitutions.

$\begin{aligned} I & = \int_2^3 (x+1)\sqrt{\frac {x+2}{x-2}} dx & \small \blue{\text{Let }u^2 = x-2 \implies 2u\ du = dx} \\ & = \int_0^1 2(u^2+3) \sqrt{u^2 + 4} \ du & \small \blue{\text{Let }\sinh t = \frac u2 \implies \cosh t \ dt = \frac {du}2} \\ & = \int_0^{\ln \varphi} 8(4\sinh^2 t + 3)\cosh^2 t \ dt & \small \blue{\text{and }\sinh^{-1} \frac 12 = \ln \left(\frac 12 + \frac {\sqrt 5}2 \right) = \ln \varphi,} \\ & = \int_0^{\ln \varphi} \left(8\sinh^2 2t + 12\cosh 2t + 12 \right) dt & \small \blue{\text{where }\varphi \text{ denotes the golden ratio.}} \\ & = \int_0^{\ln \varphi} \left(4\cosh 4t + 12\cosh 2t + 8 \right) dt \\ & = \sinh 4t + 6 \sinh 2t + 8t \ \bigg|_0^{\ln \varphi} \\ & = \frac 12 \left(\varphi^4 - \frac 1{\varphi^4} \right) + 3 \left(\varphi^2 - \frac 1{\varphi^2} \right) + 8 \ln \varphi \\ & = \frac 12 \left(3 \varphi + 2 - 5 + 3\varphi \right) + 3 \left(\varphi + 1 - 2 + \varphi \right) + 8 \ln \varphi \\ & = 9\varphi - \frac 92 + 8\ln \varphi = \frac {9\sqrt 5}2 + 8 \ln \varphi \approx \boxed{13.9} \end{aligned}$

---

$\begin{aligned} I & = \int_2^3 (x+1)\sqrt{\frac {x+2}{x-2}} dx & \small \blue{\text{Let }u^2 = x-2 \implies 2u\ du = dx} \\ & = \int_0^1 2(u^2+3) \sqrt{u^2 + 4} \ du & \small \blue{\text{Let }\tan \theta = \frac u2 \implies \sec^2 \theta \ d\theta = \frac {du}2} \\ & = \int_0^{\tan^{-1} \frac 12} 8(4 \tan^2 \theta + 3) \sec^3 \theta \ d\theta \\ & = \int_0^{\tan^{-1} \frac 12} 8(4 \sec^2 \theta - 1) \sec^3 \theta \ d\theta \\ & = 32 \blue{\int_0^{\tan^{-1} \frac 12} \sec^5 \theta \ d\theta} - 8 \int_0^{\tan^{-1} \frac 12} \sec^3 \theta \ d\theta & \small \blue{\text{By reduction formula (see note)}} \\ & = 32 \left(\blue{\frac {\tan \theta \sec^3 \theta}4\ \bigg|_0^{\tan^{-1} \frac 12} + \frac 34 \int_0^{\tan^{-1} \frac 12} \sec^3 \theta \ d\theta} \right) - 8 \int_0^{\tan^{-1} \frac 12} \sec^3 \theta \ d\theta \\ & = \frac {5\sqrt 5}2 + 16 \int_0^{\tan^{-1} \frac 12} \sec^3 \theta \ d \theta & \small \blue{\text{By reduction formula again}} \\ & = \frac {5\sqrt 5}2 + 8 \tan \theta \sec \theta \ \bigg|_0^{\tan^{-1} \frac 12} + 8 \int_0^{\tan^{-1} \frac 12} \sec \theta \ d \theta \\ & = \frac {5\sqrt 5}2 + 2\sqrt 5 + 8 \int_0^{\tan^{-1} \frac 12} \frac {\tan \theta \sec \theta + \sec^2 \theta}{\tan \theta + \sec \theta} d\theta \\ & = \frac {9\sqrt 5}2 + 8 \ln (\tan \theta + \sec \theta) \ \bigg|_0^{\tan^{-1} \frac 12} \\ & = \frac {9\sqrt 5}2 + 8 \ln \left(\frac 12 + \frac {\sqrt 5}2 \right) \approx \boxed{13.9} \end{aligned}$

---

Reduction formula: $\displaystyle \int \sec^n x\ dx = \frac {\tan x \sec^{n-2} x}{n-1} + \frac {n-2}{n-1} \int \sec^3 x \ dx$