Shorter way to integrate $\int \frac{x^9}{(x^2+4)^6} \, \mathrm{d}x$ $$
I=\int \frac{x^9}{(x^2+4)^6}\mathrm{d}x
$$
Yeah I know, I can substitute:
$$t=x^2+4\text{ or }2\tan\theta$$
So that:
$$I=\frac12\int\frac{(t-4)^4}{t^6}\mathrm{d}t\text{ or } I=2^{-2}\int\tan^9\theta\cos^{10}\theta\mathrm{d}\theta$$
Both of which are a long tedious way* to solve, is there any easier method?
Update: I am not asking among these two, I am asking any "substitution" except these two, which is shorter.
Edit: I am very sorry I missed the ^$6$ in question.



*

*Use Binomial Theorem for first and then divide by $t^6$, then integrate term wise.

*$\require{cancel}\cancel{\text{Use Reduction formula for second or integration by parts.}}$


 A: Hint :
Rewrite the integrand as
$$\frac{x^9}{x^2+4}=x^7-4x^5+16x^3+\frac{256x}{x^2+4}-64x.$$

Edit :
The OP was changed into
$$
\int\frac{x^9}{(x^2+4)^6}\ dx.
$$
Use $x=2\tan\theta\ \Rightarrow\ dx=2\sec^2\theta\ d\theta$.
$$
\int\frac{x^9}{(x^2+4)^6}\ dx=\frac14\int\tan^9\theta\ \cos^{10}\theta\ d\theta=\frac14\int\sin^9\theta\cos\theta\ d\theta.
$$
Now, set $t=\sin\theta\ \Rightarrow\ dt=\cos\theta\ d\theta$.
$$
\int\frac{x^9}{(x^2+4)^6}\ dx=\frac14\int t^9 dt=\frac1{40}t^{10}+C,
$$
where $\sin\theta=\dfrac x{\sqrt{x^2+4}}$. The answer is $\color{blue}{\text{D}}$.
A: You don't have to use a substitution quite yet. If you long-divide the integrand, you get $$\frac{x^9}{x^2+4}=x^7-4x^5+16x^3+\frac{256x}{x^2+4}-64x.$$Now, it is quite simple if the integrate term-by-term. Finishing things off: $$ \begin {align*} \displaystyle\int \frac {x^9}{x^2+4} \, \mathrm{d}x &= \displaystyle\int \left( \frac{x^9}{x^2+4}=x^7-4x^5+16x^3+\frac{256x}{x^2+4}-64x \right) \, \mathrm{d}x \\&= \frac {x^8}{8} - \frac {2}{3} x^6 + 4x^4 - 64x + 256 \displaystyle\int \frac {x}{x^2 + 4} \, \mathrm{d}x. \end {align*} $$The last integral can be done with an easy substitution.

Edit: So apparently the problem got changed again, but the same idea applies. You can re-write the new integrand as $$ \frac {x^9}{(x^2+4)^6} = \frac {x}{(x^2+4)^2} - \frac {16x}{(x^2+4)^3} + \frac {96x}{(x^2+4)^4} - \frac {256x}{(x^2+4)^5} + \frac {256x}{(x^2+4)^6}, $$ and use the substitution $u=x^2+4$ on each of these. 
A: $$I=\int \frac{x^9}{(x^2+4)^6}dx=\frac{1}{2}\int \frac{2x(x^2)^4}{(x^2+4)^6}dx=$$
$$x^2+4=u,x^2=u-4,2xdx=du$$
$$=\frac{1}{2}\int \frac{(u-4)^4}{u^6}du$$
A: The substitution $t=1+ 4/x^2$ seems to be useful in your case.
We have $$x^2+4 = \frac{4t}{t-1},$$
$$x^9 \frac{dx}{dt} = - \frac18 x^{12} = -\frac18\frac{4^6}{(1-t)^6},$$
and thus
$$\int  \frac{x^9}{(x^2+4)^6} dx = - \frac18 \int t^{-6} dt. $$
