15. $int\frac{dx}{(2x - 3)sqrt{5 - 12x + 4x^{2}}}$

Explanation:

Step1: Complete the square for the quadratic in the square - root

First, rewrite $5 - 12x+4x^{2}$ as $4x^{2}-12x + 5=4\left(x^{2}-3x\right)+5$. Completing the square inside the parentheses: $x^{2}-3x=\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}$. So $4x^{2}-12x + 5=4\left[\left(x - \frac{3}{2}\right)^{2}-\frac{9}{4}\right]+5=4\left(x-\frac{3}{2}\right)^{2}-9 + 5=4\left(x-\frac{3}{2}\right)^{2}-4$. Let $u=x-\frac{3}{2}$, then $x=u+\frac{3}{2}$ and $dx = du$. The integral becomes $\int\frac{du}{(2u)\sqrt{4u^{2}-4}}=\frac{1}{4}\int\frac{du}{u\sqrt{u^{2}-1}}$.

Step2: Use the substitution $u=\sec\theta$

If $u = \sec\theta$, then $du=\sec\theta\tan\theta d\theta$. And $\sqrt{u^{2}-1}=\tan\theta$. The integral $\frac{1}{4}\int\frac{du}{u\sqrt{u^{2}-1}}$ becomes $\frac{1}{4}\int\frac{\sec\theta\tan\theta d\theta}{\sec\theta\tan\theta}=\frac{1}{4}\int d\theta$.

Step3: Integrate

$\frac{1}{4}\int d\theta=\frac{1}{4}\theta + C$.

Step4: Back - substitute

Since $u=\sec\theta$, then $\theta=\text{arcsec}(u)$. And since $u=x - \frac{3}{2}$, the final result is $\frac{1}{4}\text{arcsec}\left(x-\frac{3}{2}\right)+C$.

Answer:

$\frac{1}{4}\text{arcsec}\left(x-\frac{3}{2}\right)+C$