Solving $\int \frac{x^3}{(4x^2 + 9)^\frac{3}{2}} dx$ The integral I must solve is this:
$$\begin{equation} \int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx \tag{1}\end{equation}$$
Now I know that I'm supposed to convert the denominator into some form of $\sqrt{a^2 + x^2}$ in order to apply the substitution $x = a \ tan \ \theta$. Now I can change the form of the denominator to become easy enough to substitute as follows:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +x^2\bigg)\bigg) ^\frac{3}{2} $$
Which makes it clear that $x$ needs to be substituted as  $x = \frac{3}{2} \tan \theta $, and $dx = \frac{3}{2} \sec^2 \theta $
for later use. At this point I can represent $(1)$ in terms of my substituted trignometric function. The only problem comes with the denominator where I get stuck on the power. Here is how I went about solving it:
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\left(\frac{3}{2} \tan \theta\right)^2 \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \bigg (\ \frac{9}{4}\ +\frac{9}{4} \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$\bigg (4 \frac{9}{4}\bigg (\ 1 + \tan^2 \theta) \bigg)\bigg) ^\frac{3}{2} $$
$$ 9^\frac{3}{2}\bigg( \ 1 + \tan^2 \theta \bigg) ^\frac{3}{2} $$
$$ 27\ ( \sec^2 \theta ) ^\frac{3}{2} $$
Now I have no idea how to evaluate this power of $sec$. The author says that it changes into $sec^3 \theta$ but I just can't fathom how that would go about. If what I understand is correct, the power it is raised to would be added to it's own making it $ \sec^\frac{7}{2} \theta$. My question is that how exactly is my reasoning wrong here?
 A: So we want to solve
$${\int \frac{x^3}{\left(4x^2 + 9\right)^{\frac{3}{2}}}dx}$$
The first thing I do when I see an integral like this is ask if there are any simple substitutions I can make. Well, notice if I make the substitution ${u=4x^2 + 9}$, then...
$${\Rightarrow \int \frac{x^3}{u^{\frac{3}{2}}}\times \frac{1}{8x} dx=\int \frac{1}{8}\frac{x^2}{u^{\frac{3}{2}}} du=\frac{1}{8}\int \frac{\frac{u-9}{4}}{u^{\frac{3}{2}}} du}$$
And this equals
$${=\frac{1}{32} \int \frac{1}{\sqrt{u}} - \frac{9}{u^{\frac{3}{2}}}du}$$
Can you take it from here?
A: $$\int \frac{x^3}{\left(4x^2+9\right)^{\frac{3}{2}}}dx$$
if $4x^2+9=t^2$ as in the answer of the user @Anton Vrdoljak you have:
$$=\int \frac{t-9}{32t^{\frac{3}{2}}}\ dt=\frac{1}{32}\cdot \int \left(\frac{1}{t^{\frac{1}{2}}}-\frac{9}{t^{\frac{3}{2}}}\right)dt=\frac{1}{32}\left(2t^{\frac{1}{2}}-\left(-\frac{18}{t^{\frac{1}{2}}}\right)\right)+k, \quad k\in\Bbb R$$
And at the end:
$$=\frac{1}{32}\left(2\left(4x^2+9\right)^{\frac{1}{2}}-\left(-\frac{18}{\left(4x^2+9\right)^{\frac{1}{2}}}\right)\right)+k, \quad k\in\Bbb R$$
$$=\frac{2x^2+9}{8\left(4x^2+9\right)^{\frac{1}{2}}}+k, \quad k\in\Bbb R$$
A: Hint: Use substitution $\space 9+4x^2 = t^2. \space$ New integral shall be the integral of rational function...
A: Apparently I can evaluate integrals but not powers. To the ones who stumble upon this, when something is raised to a power, it is multiplied to that not added to it, which was the mistake that I was making. So for my problem above
$$(\sec^{2} \theta)^{\frac{3}{2}} = ((\sec \theta)^{2})^{\frac{3}{2}} = (\sec \theta)^{2 \times \frac{3}{2}} = \sec^3 \theta  $$
A: The OP correctly identified the power rule for exponents
$$(b^n)^m= b^{n\times m}=b^{nm}.$$
Therefore for $b=\sec\theta,n=2,$ and $m=3/2$,
$$27\ ( \sec^2 \theta ) ^\frac{3}{2}=27(\sec\theta)^{2\times\frac32}=27(\sec^3\theta).$$
To finish the problem, the OP can apply the substitution $x = \frac{3}{2} \tan \theta \implies dx = \frac{3}{2} \sec^2 \theta \,d\theta.$
$$\int \frac{x^3 }{(4x^2 + 9)^\frac{3}{2}} dx=\int\frac{\frac{27}{8}\tan^3\theta}{27\sec^3\theta}\frac{3}{2} \sec^2 \theta \,d\theta$$
$$=\frac{3}{16}\int\frac{\tan^3\theta}{\sec^3\theta}\sec^2 \theta \,d\theta$$
$$=\frac{3}{16}\int\frac{\tan^3\theta}{\sec\theta} \,d\theta$$
$$=\frac{3}{16}\int\frac{(\sec^2\theta-1)\tan\theta}{\sec\theta} \,d\theta$$
$$=\frac{3}{16}\left(\int\,d(\sec \theta)+\int\,d(\cos\theta)\right)$$
$$=\frac{3}{16}\left(\sec \theta+\cos\theta\right)+C.$$
The substitution $x = \frac{3}{2} \tan \theta \implies \sqrt{4x^2+9}=3\sec\theta$ which forms
$$=\frac{3}{16}\left(\frac{\sqrt{4x^2+9}}{3}+\frac{3}{\sqrt{4x^2+9}}\right)+C$$
$$=\frac{3}{16}\left(\frac{4x^2+18}{3\sqrt{4x^2+9}}\right)+C$$
$$=\boxed{\frac{2x^2+9}{8\sqrt{4x^2+9}}+C}$$

As an alternative approach, I observed that
$$x^3=\frac{x(4x^2+9)-9x}{4},$$
therefore
$$\int \frac{x^3}{(4x^2+9)^{3/2}}\,dx=\frac{1}{4}\int \frac{x(4x^2+9)-9x}{(4x^2+9)^{3/2}}\,dx$$
$$=\frac{1}{4}\int \frac{x}{\sqrt{4x^2+9}}\,dx-\frac{9}{4}\int\frac{x}{(4x^2+9)^{3/2}}\,dx$$
Substituting $u=4x^2+9 \implies du=8x\,dx$ on both integrals forms
$$=\frac{1}{32}\int\frac{1}{\sqrt{u}}\,du-\frac{9}{32}\int\frac{1}{u^{3/2}}\,du$$
$$=\frac{1}{16}\sqrt{u}+\frac{9}{16\sqrt{u}}+C$$
$$=\frac{1}{16}\left(\frac{u+9}{\sqrt{u}}\right)+C$$
$$=\frac{1}{16}\left(\frac{4x^2+18}{\sqrt{4x^2+9}}\right)+C$$
$$=\boxed{\frac{2x^2+9}{8\sqrt{4x^2+9}}+C}$$
A: You can easily integrate with suitable substitution as follows
Let $2x=3\tan\theta\implies dx=\frac{3}{2}\sec^2\theta\ d\theta$
$$\int \frac{x^3}{(4x^2+9)^{3/2}}dx=\int \frac{(\frac32\tan\theta)^3}{(9\tan^2\theta+9)^{3/2}}\ \frac{3}{2}\sec^2\theta\ d\theta$$
$$=\left(\frac{3}{2}\right)^4\frac{1}{3^3}\int \frac{\tan^3\theta}{\sec^3\theta}\sec^2\theta\ d\theta$$
$$=\frac{3}{16}\int \frac{\tan\theta(\sec^2\theta-1)}{\sec\theta}\ d\theta$$
$$=\frac{3}{16}\int (\sec\theta\tan\theta-\sin\theta)\ d\theta$$
$$=\frac{3}{16}(\sec\theta+\cos\theta)+C$$
substitute back to $x$,
$$=\frac{3}{16}\left(\frac{\sqrt{4x^2+9}}{3}+\frac{3}{\sqrt{4x^2+9}}\right)+C$$
$$=\color{blue}{\frac{2x^2+9}{8\sqrt{4x^2+9}}}+C$$
A: Here is another easier way to integrate as follows
$$\int \frac{x^3}{(4x^2+9)^{3/2}}dx=\int \frac14\frac{x(4x^2+9)-9x}{(4x^2+9)^{3/2}}dx$$
$$=\frac14\int \frac{x}{\sqrt{4x^2+9}}dx-\frac14\int \frac{9x}{(4x^2+9)^{3/2}}dx$$
$$=\frac1{32}\int \frac{d(4x^2+9)}{\sqrt{4x^2+9}}-\frac9{32}\int \frac{d(4x^2+9)}{(4x^2+9)^{3/2}}$$
$$=\frac1{32}2\sqrt{4x^2+9}-\frac9{32}\frac{-2}{\sqrt{4x^2+9}}+C$$
$$=\bbox[15px,#ffd,border:1px solid green ]{\frac{2x^2+9}{8\sqrt{4x^2+9}}+C}$$
