91Ó°ÊÓ

First, notice that the denominator can be rewritten as \((3+(-x+4))^2=(3+(-x+4))^2\). This resembles a Pythagorean identity, specifically \((\sqrt{x^2+a^2)}\), which leads us to use a trigonometric substitution of the form \(x=a\tan(\theta)\). Since we have \(a^2=3^2\) and \((-x+4)^2\) inside the square root, we can use the substitution \(x=4-3\tan(\theta)\). Now we will compute the differential, \(dx\). $$dx=-3\sec^2(\theta)\ d\theta$$

Now we will rewrite the integral with the substitution and differential. $$\int \frac{(4-3\tan(\theta))^2-8 (4-3\tan(\theta))+16}{\left(9+8 (4-3\tan(\theta))-(4-3\tan(\theta))^2\right)^{3 / 2}} (-3\sec^2(\theta))d\theta$$ Now, simplify the integral to a more manageable form by expanding, combining, and canceling terms. $$-\int \frac{9\tan^2(\theta)-24\tan(\theta)+16}{(9-(9\tan^2(\theta)))^{3 / 2}}(-3\sec^2(\theta) )d\theta$$ $$=-3\int \frac{\tan^2(\theta)-8\tan(\theta) + 16}{(3 \sec(\theta))^{3}}\sec^2(\theta) d\theta$$ $$=-\frac{1}{27}\int \frac{\tan^2(\theta)-8\tan(\theta)+16}{\sec^2(\theta)}d\theta$$

Since \(\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\) and \(\sec(\theta)=\frac{1}{\cos(\theta)}\), let's rewrite the integral in terms of sin and cos and combine expressions inside the integral. $$-\frac{1}{27}\int \frac{(\frac{\sin^2(\theta)}{\cos^2(\theta)})-8(\frac{\sin(\theta)}{\cos(\theta)})+16}{\frac{1}{\cos^2(\theta)}}d\theta$$ $$-\frac{1}{27}\int \frac{\sin^2(\theta)}{\cos^2(\theta)}-8\frac{\sin(\theta)}{\cos(\theta)}+16\ d\theta$$

Now we need to convert back to x using the substitution. Recall that \(x=4-3\tan(\theta)\). Using the right triangle representation of the trig functions, we can see that opposite = \(x-4\), adjacent = 3, and hypotenuse = \(\sqrt{(-x+8)^{2}+3^2}\). So, \(\sin(\theta)=\frac{(x-4)}{\sqrt{(-x+8)^{2}+3^2}}\) and \(\cos(\theta)=\frac{3}{\sqrt{(-x+8)^{2}+3^2}}\). Now we substitute these expressions into the integral. $$-\frac{1}{27}\int \frac{((x-4)^2)/3^2}{((x-4)^2)/3^2} - 8\frac{(x-4)/3}{((-x+8)^2+3^2)}+16\,dx$$ Now, integrate each term separately. $$-\frac{1}{27} \left[\int 1\,dx - 8 \int \frac{x-4}{((-x+8)^2+3^2)}\,dx + 16 \int \frac{1}{((-x+8)^2+3^2)} \,dx\right]$$ The integrals of the three terms are: 1. \(\int 1 dx = x+C_1\) 2. \(\int \frac{x-4}{((-x+8)^2+3^2)} dx\) could be integrated using substitution or by simple inspection: it's an odd function so it returns 0. 3. \(\int \frac{1}{((-x+8)^2+3^2)} dx\) can be integrated using an arctan substitution. It results in \(\frac{1}{3} \arctan{\left(\frac{x-4}{3}\right)} + C_2\). Putting all three integrals together, we get: $$-\frac{1}{27}(x - 8\cdot0 +\frac{16}{3}\arctan{\left(\frac{x-4}{3}\right)} +C)$$

The final solution for the given integral is: $$-\frac{1}{27}(x +\frac{16}{3}\arctan{\left(\frac{x-4}{3}\right)} +C)$$