Carry out each specified substitution

$u=2-3x$ gives $du/dx=-3$, $dx/du=-1/3$ and $x=(2-u)/3$. Therefore the integral becomes $\int\frac{2-u}{3}u^{11}(-\frac13)\,du=-\frac19\int(2u^{11}-u^{12})\,du$.

The result is $-\frac19(\frac{2u^{12}}{12}-\frac{u^{13}}{13})+C=-\frac{u^{12}}{54}+\frac{u^{13}}{117}+C=-\frac{(2-3x)^{12}}{54}+\frac{(2-3x)^{13}}{117}+C$. These are the source four steps: differentiate the substitution, replace every x and dx, integrate in u, then return to x.

For $x>2$, $u^2=x-2$ gives $2u\,du/dx=1$, $dx/du=2u$ and $x=u^2+2$. Hence $\int\frac{x+1}{x\sqrt{x-2}}\,dx=\int\frac{u^2+3}{(u^2+2)u}(2u)\,du=2\int\frac{u^2+3}{u^2+2}\,du$.

The source asks how to handle the improper rational function; the Teacher supplies long division. Since $(u^2+3)/(u^2+2)=1+1/(u^2+2)$, the result is $2[u+\frac1{\sqrt2}\tan^{-1}(u/\sqrt2)]+C=2\sqrt{x-2}+\sqrt2\tan^{-1}(\sqrt{x-2}/\sqrt2)+C$.

$x=2\tan\theta$ gives $dx=2\sec^2\theta\,d\theta$. Since $\cos\theta>0$ on the stated interval, $(4\tan^2\theta+4)^{3/2}=4^{3/2}(\sec^2\theta)^{3/2}=8\sec^3\theta$. The integral is $\int\frac{2\sec^2\theta}{8\sec^3\theta}\,d\theta=\frac14\int\cos\theta\,d\theta=\frac14\sin\theta+C$.

The Teacher triangle for $x>0$ has opposite side x, adjacent side 2 and hypotenuse $\sqrt{x^2+4}$. Thus $\sin\theta=x/\sqrt{x^2+4}$ and the result is $x/[4\sqrt{x^2+4}]+C$. Signed principal-angle identities give the same formula for every real x. The source recommends a quick right-triangle sketch instead of leaving $\sin(\tan^{-1}(x/2))$ unevaluated.

$dx=3\cos\theta\,d\theta$ and $\sqrt{9-x^2}=3|\cos\theta|=3\cos\theta$ on the stated interval. On $-\pi/2<\theta<\pi/2$, $\theta\ne0$, the transformed integral is $\int\frac{3\cos\theta}{9\sin^2\theta}(3\cos\theta)\,d\theta=\int\cot^2\theta\,d\theta=\int(\csc^2\theta-1)\,d\theta=-\cot\theta-\theta+C$.

The source triangle for $x>0$ has hypotenuse 3, opposite side x and adjacent side $\sqrt{9-x^2}$. Therefore the antiderivative is $-\sqrt{9-x^2}/x-\sin^{-1}(x/3)+C$, initially on each interval $-3<x<0$ and $0<x<3$. Signed ratios extend the working to negative x. The printed closed angle range includes endpoints; endpoint values can be taken by one-sided limits, while the substitution calculation uses open intervals and x=0 remains excluded.