Euler's substitution might just work!

Putting $x = 3\sinh u$ gives (note that $\sinh^{-1}\tfrac43 = \ln3$ ): $\begin{aligned} \int_0^4 \frac{\sqrt{x^2+9}\big(x + \sqrt{x^2+9}\big)}{\sqrt{x^2+9} - x}\,dx & = \; 9\int_0^{\ln3} \cosh^2u \,e^{2u}\,du \; =\; \tfrac94\int_0^{\ln3} (e^{4u} + 2e^{2u} + 1)\,du \\ & = \tfrac94\Big[ \tfrac14e^{4u} + e^{2u} + u\Big]_0^{\ln3} \; = \; 63 + \tfrac94\ln3 \end{aligned}$

First we set Euler substitution $\sqrt{x^2 + 9} = x + t$ . This leads to:

$x^2 + 9 = x^2 + 2xt + t^2$

$\large x = \frac{9-t^2}{2t}$

$\large dx = - \frac{9 + t^2}{2t^2} dt$

$\large \sqrt{x^2 + 9} = \sqrt{ \frac{81-18t^2 + t^4}{4t^2} +9} = \frac{9+t^2}{2t}$

So:

$\large \displaystyle \large \int_0^4 \frac{\sqrt{x^2 + 9}(x + \sqrt{x^2 + 9})}{\sqrt{x^2 + 9} - x} dx$

Becomes:

$\large \displaystyle \large \int_3^1 \frac{\frac{9+t^2}{2t}(\frac{9-t^2}{2t} + \frac{9+t^2}{2t})}{\frac{9+t^2}{2t} - \frac{9-t^2}{2t}} \cdot - \frac{9 + t^2}{2t^2} dt$

$= \large \displaystyle \large \int_1^3 \frac{\frac{9+t^2}{2t} \cdot \frac{18}{2t}} {t} \cdot \frac{9 + t^2}{2t^2} dt$

$= \large \displaystyle \large \frac94 \int_1^3 \frac{(9+t^2)^2}{t^5} dt$

$= \large \displaystyle \large \frac94 \int_1^3 \Big ( \frac{1}{t} - \frac{18}{t^3} + \frac{81}{t^5} \Big) dt$

$= \large \displaystyle \large \frac94 \Big ( ln(t) \Big |_1^3 - \frac{9}{t^2} \Big |_1^3 - \frac{81}{4t^4} \Big |_1^3 \Big)$

$= \large \displaystyle \large \frac94 \Big (ln(3) - (1 - 9) - (\frac{1}{4} - \frac{81}{4} ) \Big )$

$= \large \displaystyle \large \frac94 \Big (ln(3) + 28 \Big )$

$= \large \displaystyle \large 63 + \frac94 ln(3)$

$\color{#3D99F6} \approx \large \displaystyle \large \fbox{65.472}$

$\begin{aligned} I & = \int_0^4 \frac {\sqrt{x^2+9}\left(x + \sqrt{x^2+9} \right)}{\sqrt{x^2+9}-x} dx \\ & = \int_0^4 \frac {\sqrt{x^2+9}\left(x + \sqrt{x^2+9} \right)^2}{\left(\sqrt{x^2+9}-x\right)\left(\sqrt{x^2+9} + x\right)} dx \\ & = \frac 19 \int_0^4 \sqrt{x^2+9}\left(x + \sqrt{x^2+9} \right)^2 dx \qquad \small \color{#3D99F6} \text{Let } x = 3 \tan \theta, \ dx = 3 \sec^2 \theta \ d\theta \\ & = \frac 19 \int_0^{\tan^{-1} \frac 43} 3\sec \theta \left(3 \tan \theta + 3 \sec \theta \right)^2 \cdot 3 \sec^2 \theta \ d \theta \\ & = 9 \int_0^{\tan^{-1} \frac 43} \sec^3 \theta \left( \tan \theta + \sec \theta \right)^2 \ d \theta \\ & = 9 \int_0^{\tan^{-1} \frac 43} \sec^3 \theta \tan^2 \theta + 2 \sec^4 \theta \tan \theta + \sec^5 \theta \ d \theta \\ & = 9 \int_0^{\tan^{-1} \frac 43} \sec^5 \theta - \sec^3 \theta + 2 \sec^4 \theta \tan \theta + \sec^5 \theta \ d \theta \\ & = 9 \int_0^{\tan^{-1} \frac 43} 2 \sec^5 \theta - \sec^3 \theta + 2 \sec^4 \theta \tan \theta \ d \theta \\ & = 18 {\color{#3D99F6}\int_0^{\tan^{-1} \frac 43} \sec^5 \theta \ d\theta} - 9 {\color{#3D99F6}\int_0^{\tan^{-1} \frac 43} \sec^3 \theta \ d \theta} + 18 \int_0^{\tan^{-1} \frac 43} \sec^4 \theta \tan \theta \ d \theta \qquad \small \color{#3D99F6} \text{See: reduction formula} \\ & = 18 \left({\color{#3D99F6} \frac 34 \int_0^{\tan^{-1} \frac 43} \sec^3 \theta \ d\theta + \frac {\sec^3 \theta \tan \theta}4 \bigg|_0^{\tan^{-1} \frac 43}}\right) - 9 \left( {\color{#3D99F6} \frac 12 \int_0^{\tan^{-1} \frac 43} \sec \theta \ d\theta + \frac {\sec \theta \tan \theta}2 \bigg|_0^{\tan^{-1} \frac 43}}\right) + 18 \int_1^\frac 53 \sec^3 \theta \ d \sec \theta \\ & = 18 \left(\frac 34 \left(\frac 12 \int_0^{\tan^{-1} \frac 43} \sec \theta \ d\theta + \frac {10}9 \right) + \frac {125}{81} \right) - 9 \left(\frac 12 \int_0^{\tan^{-1} \frac 43} \sec \theta \ d\theta + \frac {10}9 \right) + 18 \left(\frac {5^4}{3^4 4} - \frac 14\right) \\ & = \frac 94 \int_0^{\tan^{-1} \frac 43} \sec \theta \ d\theta + 63 \\ & = \frac 94 \cdot \ln (\tan \theta + \sec \theta) \bigg|_0^{\tan^{-1} \frac 43} + 63 \\ & = \frac {9\ln 3}4 + 63 \approx \boxed{65.472} \end{aligned}$

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