3. 0 / 2.08 points evaluate the integral. (remember the constant of integration.) (int\frac{x^{2}}{(15 + 6x…

Explanation:

Step1: Complete the square for the denominator

First, rewrite $15 + 6x-9x^{2}$ as $-9(x^{2}-\frac{2}{3}x - \frac{5}{3})$. Completing the square inside the parentheses: $x^{2}-\frac{2}{3}x - \frac{5}{3}=(x - \frac{1}{3})^{2}-\frac{1}{9}-\frac{5}{3}=(x - \frac{1}{3})^{2}-\frac{1 + 15}{9}=(x - \frac{1}{3})^{2}-\frac{16}{9}$. So, $15 + 6x-9x^{2}=16-9(x - \frac{1}{3})^{2}$. Let $u=x-\frac{1}{3}$, then $x = u+\frac{1}{3}$ and $dx=du$. The integral becomes $\int\frac{(u + \frac{1}{3})^{2}}{(16-9u^{2})^{\frac{3}{2}}}du=\int\frac{u^{2}+\frac{2}{3}u+\frac{1}{9}}{(16 - 9u^{2})^{\frac{3}{2}}}du=\int\frac{u^{2}}{(16 - 9u^{2})^{\frac{3}{2}}}du+\frac{2}{3}\int\frac{u}{(16 - 9u^{2})^{\frac{3}{2}}}du+\frac{1}{9}\int\frac{1}{(16 - 9u^{2})^{\frac{3}{2}}}du$.

Step2: Evaluate $\frac{2}{3}\int\frac{u}{(16 - 9u^{2})^{\frac{3}{2}}}du$

Let $t = 16-9u^{2}$, then $dt=-18udu$. So, $\frac{2}{3}\int\frac{u}{(16 - 9u^{2})^{\frac{3}{2}}}du=\frac{2}{3}\times(-\frac{1}{18})\int t^{-\frac{3}{2}}dt=-\frac{1}{27}\times(- 2)t^{-\frac{1}{2}}+C_1=\frac{2}{27(16 - 9u^{2})^{\frac{1}{2}}}+C_1$.

Step3: Evaluate $\int\frac{u^{2}}{(16 - 9u^{2})^{\frac{3}{2}}}du$ and $\frac{1}{9}\int\frac{1}{(16 - 9u^{2})^{\frac{3}{2}}}du$

Use the trigonometric - substitution $u=\frac{4}{3}\sin\theta$, then $du=\frac{4}{3}\cos\theta d\theta$ and $16 - 9u^{2}=16 - 16\sin^{2}\theta = 16\cos^{2}\theta$. For $\int\frac{1}{(16 - 9u^{2})^{\frac{3}{2}}}du=\int\frac{1}{(16\cos^{2}\theta)^{\frac{3}{2}}}\times\frac{4}{3}\cos\theta d\theta=\frac{1}{48}\int\frac{1}{\cos^{2}\theta}d\theta=\frac{1}{48}\tan\theta+C_2$. Since $u = \frac{4}{3}\sin\theta$, $\sin\theta=\frac{3u}{4}$, $\cos\theta=\frac{\sqrt{16 - 9u^{2}}}{4}$, $\tan\theta=\frac{3u}{\sqrt{16 - 9u^{2}}}$, so $\frac{1}{9}\int\frac{1}{(16 - 9u^{2})^{\frac{3}{2}}}du=\frac{u}{48\sqrt{16 - 9u^{2}}}+C_3$. For $\int\frac{u^{2}}{(16 - 9u^{2})^{\frac{3}{2}}}du=\int\frac{\frac{16}{9}\sin^{2}\theta}{(16\cos^{2}\theta)^{\frac{3}{2}}}\times\frac{4}{3}\cos\theta d\theta=\frac{4}{27}\int\frac{\sin^{2}\theta}{\cos^{2}\theta}d\theta=\frac{4}{27}\int\tan^{2}\theta d\theta=\frac{4}{27}\int(\sec^{2}\theta - 1)d\theta=\frac{4}{27}(\tan\theta-\theta)+C_4$. Substituting back $\theta=\arcsin(\frac{3u}{4})$, $\tan\theta=\frac{3u}{\sqrt{16 - 9u^{2}}}$, we get $\int\frac{u^{2}}{(16 - 9u^{2})^{\frac{3}{2}}}du=\frac{u}{9\sqrt{16 - 9u^{2}}}-\frac{4}{27}\arcsin(\frac{3u}{4})+C_4$.

Step4: Combine the results and substitute back $u=x-\frac{1}{3}$

Combining the results of the three integrals and substituting $u=x - \frac{1}{3}$ back: [ \begin{align*} &\int\frac{x^{2}}{(15 + 6x-9x^{2})^{\frac{3}{2}}}dx\ =&\left(\frac{(x-\frac{1}{3})}{9\sqrt{16-9(x - \frac{1}{3})^{2}}}-\frac{4}{27}\arcsin\left(\frac{3(x-\frac{1}{3})}{4}\right)\right)+\frac{2}{27\sqrt{16-9(x - \frac{1}{3})^{2}}}+\frac{(x-\frac{1}{3})}{48\sqrt{16-9(x - \frac{1}{3})^{2}}}+C\ =&\frac{16(x-\frac{1}{3}) + 8+3(x-\frac{1}{3})}{144\sqrt{16-9(x - \frac{1}{3})^{2}}}-\frac{4}{27}\arcsin\left(\frac{3x - 1}{4}\right)+C\ =&\frac{16x-16 + 8+3x - 1}{144\sqrt{16-9(x^{2}-\frac{2}{3}x+\frac{1}{9})}}-\frac{4}{27}\arcsin\left(\frac{3x - 1}{4}\right)+C\ =&\frac{19x - 9}{144\sqrt{15 + 6x-9x^{2}}}-\frac{4}{27}\arcsin\left(\frac{3x - 1}{4}\right)+C \end{align*} ]

Answer:

$\frac{19x - 9}{144\sqrt{15 + 6x-9x^{2}}}-\frac{4}{27}\arcsin\left(\frac{3x - 1}{4}\right)+C$