Simple integral $\int \frac{1}{{(1+x^2)^3}} \ dx $ How would you find $$\int \frac{1}{{(1+x^2)^3}} \ dx \ ?$$ I think it calls for some trig substitution, but I tried and it didn't work (or at least I couldn't see it how would work).
 A: We want to know $\int \frac{1}{(\color{red}{1+x^2})^3} \color{green}{dx}$.
Let $x=\tan(\theta) \iff \theta=\arctan(x)$.
Then $\frac{dx}{d\theta}=\sec^2(\theta) \iff \color{green}{dx=\sec^2(\theta)d\theta}$.
Now you've got $\int \frac{1}{[\color{red}{1+\tan^2(\theta)}]^3} \color{green}{\sec^2(\theta)d\theta}=\int \frac{1}{[\color{red}{\sec^2(\theta)}]^3}\color{green}{\sec^2(\theta)d\theta}=\int\frac{\color{green}{\sec^2(\theta)}}{\sec^6(\theta)}\color{green}{d\theta}= \overbrace{\int \frac{1}{\sec^4(\theta)} d\theta=\int \cos^4(\theta) d\theta}^{\textrm{remeber that sec}(\theta)=\frac{1}{\cos(\theta)}}$.
But this is $\underbrace{\int [\cos^2(\theta)]^2d\theta=\int \left\{ \frac{1}{2}[1+\cos(2\theta)] \right\}^2 d\theta}_{\textrm{using the identity} \color{purple}{\cos(2\theta) \equiv 2\cos^2(\theta)-1}}=\frac{1}{4}\int [1+\cos(2\theta)]^2 d\theta=??$
I'll leave the rest up to you. Remember that, to integrate $\cos^2(\alpha),$ use the $\color{purple}{\text{identity}}$ that I've given. Don't forget to sub in $\theta=\arctan(x)$ at the end; that's something I'm always guilty of!
By the way, as a general rule, if, ever, you've got a function of $1 \pm x^2, \sqrt{1 \pm x^2}$ (and nothing else), make it your first instinct to make a suitable trig./hyperbolic substitution.
A: Hint: Let $F(a)=\displaystyle\int\frac{dx}{a+x^2}$ , then compute $F''(1)$.
A: By $x=\sec \theta$, you'll get $$I=\int \cos^4\theta d\theta=\dfrac {1}{2^2}\int (1+\cos 2\theta)^2 d\theta$$ Hope you can do the rest yourself.
