Commpute INtegral

$\begin{aligned} I & = \int_1^5 \frac {2-x}{\left(x^2+2\right)^\frac 32} \ dx \\ & = \int_1^5 \frac {2-x}{2\sqrt 2\left(\frac {x^2}2+1\right)^\frac 32} \ dx & \small \color{#3D99F6} \text{Let } \tan \theta = \frac x{\sqrt 2} \implies \sec^2 \theta \ d\theta = \frac {dx}{\sqrt 2} \\ & = \int_{\tan^{-1}\frac 1{\sqrt 2}}^{\tan^{-1}\frac 5{\sqrt 2}} \frac {(2-\sqrt 2\tan \theta)\sec^2 \theta}{2\left(\tan^2 \theta +1\right)^\frac 32} \ d\theta \\ & = \int_{\tan^{-1}\frac 1{\sqrt 2}}^{\tan^{-1}\frac 5{\sqrt 2}} \frac {1 - \frac 1{\sqrt 2}\tan \theta}{\sec \theta} \ d\theta \\ & = \int_{\tan^{-1}\frac 1{\sqrt 2}}^{\tan^{-1}\frac 5{\sqrt 2}} \left(\cos \theta - \frac {\sin \theta}{\sqrt 2}\right) d\theta \\ & = \left[\sin \theta + \frac {\cos \theta}{\sqrt 2}\right]_{\tan^{-1}\frac 1{\sqrt 2}}^{\tan^{-1}\frac 5{\sqrt 2}} \\ & = \frac 5{3\sqrt 3} + \frac {\sqrt 2}{\sqrt 2 \cdot 3 \sqrt 3} - \frac 1{\sqrt 3} - \frac {\sqrt 2}{\sqrt 2\cdot \sqrt 3} \\ & = \frac 2{\sqrt 3} - \frac 2{\sqrt 3} \\ & = \boxed{0} \end{aligned}$