Is $x=\tan\theta$ the only way for this integral? Evaluate $$\int\frac{dx}{(1+x^2)^4}$$ Now I did solve it, but I used the mentioned substitution and after a lot of converting into double angles, I did it. But, it doesn't look like a good approach. Is there a better way? Thanks.
 A: The standard way for $I_n=\displaystyle\int\frac{\mathrm d\mkern1mu x}{(1+x^2)^n} $ is to use integration by parts, setting $$u=\frac1{(1+x^2)^n}, \quad \mathrm d\mkern1mu v=\mathrm d\mkern1mu x,$$
to obtain a recurrence relation between $I_n$ and $I_{n+1}$. This relation is initialised by $\, I_1=\arctan x$.
A: Try this trick, which will lead you to a recurrence formula
$$\int\frac{dx}{(1+x^2)^n} = \int\frac{1 + x^2 - x^2}{(1+x^2)^n} dx$$
Let me know if it's not enough for a hint.
A: Just a guess (I haven't tried it) but could you change it into partial fractions?
A: If you let $x=\tan\theta$ then
$$\begin{align}\int\frac{dx}{(x^2+1)^4}&=\int\cos^6\theta\,d\theta\\
&=\frac1{32}\int\left(\cos6\theta+6\cos4\theta+15\cos2\theta+10\right)d\theta\\
&=\frac1{32}\left(\frac16\sin6\theta+\frac32\sin4\theta+\frac{15}2\sin2\theta+10\theta\right)+C\\
&=\left(\frac16\left(32\sin^5\theta-32\sin^3\theta+6\sin\theta\right)\cos\theta\right.\\
&\left.+\frac32\left(4\sin\theta-8\sin^3\theta\right)\cos\theta+\frac{15}2\left(2\sin\theta\right)\cos\theta+10\theta\right)+C\\
&=\frac1{32}\left[\left(\frac{16}3\frac{x^5}{(x^2+1)^{5/2}}-\frac{52}3\frac{x^3}{(x^2+1)^{3/2}}+22\frac x{(x^2+1)^{1/2}}\right)\frac1{(x^2+1)^{1/2}}+10\tan^{-1}x\right]+C\\
&=\frac16\frac{x^5}{(x^2+1)^3}-\frac{13}{24}\frac{x^3}{(x^2+1)^2}+\frac{11}{16}\frac x{(x^2+1)}+\frac5{16}\tan^{-1}x+C\end{align}$$
In the above I expanded $\cos^6\theta=\left(\frac{e^{i\theta}+e^{-i\theta}}2\right)^6$ by the binomial theorem and then used ran the Chebyshev polynomials of the second kind out to
$$\begin{array}{rl}\sin(2\theta)=&2\sin\theta\cos\theta\\
\sin(3\theta)=&3\sin\theta-4\sin^3\theta\\
\sin(4\theta)=&4\sin\theta\cos\theta-8\sin^3\theta\cos\theta\\
\sin(5\theta)=&16\sin^5\theta-20\sin^3\theta+5\sin\theta\\
\sin(6\theta)=&32\sin^5\theta\cos\theta-32\sin^3\theta\cos\theta+6\sin\theta\cos\theta\end{array}$$
Then since $x=\tan\theta$, $\cos\theta=\frac1{\sqrt{x^2+1}}$, and $\sin\theta=\frac x{\sqrt{x^2+1}}$. Since the result was fairly complicated, I verified it with numerical quadrature.
