삼각치환

Completing the square, we see $4x^2-12x+8 = (2x-3)^2-1$.

$$\begin{aligned} \int \dfrac{1}{(2x-3)^3\sqrt{4x^2-12x+8}}~\mathrm{d}x&= \int \dfrac{1}{(2x-3)^3\sqrt{(2x-3)^2-1}}~\mathrm{d}x \\ \text{As $x>2$, we have $2x-3>1$. We use the substitution $2x-3 = \sec \theta$ with $0\le\theta<\frac{\pi}{2}$. So $2\,\mathrm{d}x=\sec\theta\tan\theta~\mathrm{d}\theta$ and $\sqrt{(2x-3)^2-1}=\sqrt{\sec^2\theta-1}=\sqrt{\tan^2\theta}=\tan\theta$.} \\ &=\frac{1}{2}\int\frac{1}{\sec^3\theta\sqrt{\sec^2\theta-1}}\sec\theta\tan\theta~\mathrm{d}\theta\\ &=\frac{1}{2}\int\frac{1}{\sec^3\theta\tan\theta}\sec\theta\tan\theta~\mathrm{d}\theta \\&=\frac{1}{2}\int\frac{1}{\sec^2\theta}~\mathrm{d}\theta \\&=\frac{1}{2}\int{\cos^2\theta}~\mathrm{d}\theta \\&=\frac{1}{4}\int{\left(1+\cos(2\theta)\right)}~\mathrm{d}\theta \\&=\frac14\left(\theta + \frac{1}{2}\sin(2\theta)\right)+C \\&=\frac14\left(\theta + \sin\theta\cos\theta\right)+C \\ \smash{  }\,&=\frac{1}{4}\left(\arccos\left(\frac{1}{2x-3}\right) + \frac{\sqrt{4x^2-12x+8}}{(2x-3)^2}\right)+C \end{aligned}$$

Since $2x-3=\sec\theta$, we know $\cos\theta = \frac{1}{2x-3}$ and $\theta = \arccos\left(\frac{1}{2x-3}\right)$. (Equivalently, $\theta = \mathop{\mathrm{arcsec}}(2x-3)$.) To find $\sin\theta$, we draw a right triangle with adjacent side of length 1, and hypotenuse of length $2x-3$. By the Pythagorean Theorem, the opposite side has length $\sqrt{4x^2-12x+8}$.