<s>[INST] How to calculate the volume obtained by rotating the following around the x axis? The original problem is: 
"Find the volume of the solid obtained by rotating about the x axis the region enclosed by the curves $y = \frac{9}{x^2 + 9},y=0,x=0,\,$and $x = 3$"
I set up the following integral $$81\pi\int_0^3\frac{1}{(x^2 + 9)^2}dx$$ using the cylinder method (I believe it's called like that) and when I calculated it using a computer I obtained the correct answer but I have been having difficulties in solving it manually.  I tried the shell method as well and I didn't see it any easier to solve but I may be wrong of course. 
 [/INST] We want to find 
$$\int \frac{dx}{(x^2+9)^2}.$$
Let $x=3u$. Apart from a constant factor, we end up with
$$\int\frac{du}{(u^2+1)^2}.$$
Let $u=\tan t$ (we could have gone more directly, by letting $x=3\tan t$).  Then $du=\sec^2 t\,dt$, and we end up with
$$\int \frac{\sec^2 t}{\sec^4 t}\,dt.$$
This is the familiar integral
$$\int \cos^2 t\,dt.$$
A common approach to this is to use the fact that $\cos 2t=2\cos^2 t-1$.
e can save time if we make the substitutions on the definite integral, that is, substitute for the endpoints. 
Another way: Consider $\displaystyle\int \frac{dx}{x^2+9}$. Attack this by integration by parts, letting $du=dx$ and $v=\frac{1}{9+x^2}$. Then we can take $u=x$. Also, $dv=\frac{-2x\,dx}{(x^2+9)^2}$. Thus
$$\int \frac{dx}{x^2+9}=\frac{x}{x^2+9}+\int\frac{2x^2\,dx}{(x^2+9)^2}.$$
But $2x^2=2x^2+18-18$. Thus
$$\int\frac{dx}{x^2+9}=\frac{x}{x^2+9}+2\int\frac{dx}{x^2+9}-18\int\frac{dx}{(9+x^2)^2}.$$
We will be finished if we can find $\displaystyle\int\frac{dx}{x^2+9}$. The substitution $x=3w$ turns this into a familiar integral.
Remark: In the second approach, we obtained a reduction formula. A very similar trick expresses $\displaystyle\int \frac{dx}{(x^2+9)^{n+1}}$ in terms of 
$\displaystyle\int \frac{dx}{(x^2+9)^{n}}$
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