My hint was the \(1+x^4\) was factorisable into two real polynomials, and one way to see this is to factorise it into four linear terms involving complex numbers and to pair up linear terms until you got pairings which produced real quadratic polynomials. The other way is to complete the square: \begin{align}1+x^4=&1+2x^4+x^4-2x^2\\=&(1+x^2)^2-(\sqrt{2}x)^2\\=&(1-\sqrt{2}x+x^2)(1+\sqrt{2}x+x^2).\end{align} Now, if you're comfortable with partial fractions, you should immediately notice that: \begin{align}\frac{1}{1-\sqrt{2}x+x^2}-\frac{1}{1+\sqrt{2}x+x^2}&=\frac{2\sqrt{2}x}{1+x^4}\text{, and}\\ \frac{1}{1-\sqrt{2}x+x^2}+\frac{1}{1+\sqrt{2}x+x^2}&=\frac{2+2x^2}{1+x^4}.\end{align} Since \(1=\frac{1}{2}(2+2x^2)-\frac{x}{2\sqrt{2}}2\sqrt{2}x\), we see that: \begin{align}\frac{1}{1+x^4}=&\frac{1}{2}\left(\frac{1}{1-\sqrt{2}x+x^2}+\frac{1}{1+\sqrt{2}x+x^2}\right)\\&-\frac{x}{2\sqrt{2}}\left(\frac{1}{1-\sqrt{2}x+x^2}-\frac{1}{1+\sqrt{2}x+x^2}\right)\\=&\frac{\frac{1}{2}-\frac{x}{2\sqrt{2}}}{1-\sqrt{2}x+x^2}+\frac{\frac{1}{2}+\frac{x}{2\sqrt{2}}}{1+\sqrt{2}x+x^2}\\=& \frac{1+\sqrt{2}x}{4(1+\sqrt{2}x+x^2)}+\frac{1-\sqrt{2}x}{4(1+\sqrt{2}x+x^2)}\\&+\frac{1}{4(1+\sqrt{2}x+x^2)}+\frac{1}{4(1-\sqrt{2}x+x^2)}.\end{align} Finally, integrating the two upper-line terms into a logarithm and integrating the two lower-line terms into an inverse tan function, we get: \begin{align} \int \frac{\mathrm{d}x}{1+x^4}=&\frac{1}{4\sqrt{2}}\log|1+\sqrt{2}x+x^2|-\frac{1}{4\sqrt{2}}\log|1-\sqrt{2}x+x^2|\\&+\frac{1}{2\sqrt{2}}\mathrm{arctan}(\sqrt{2}x+1)+\frac{1}{2\sqrt{2}}\mathrm{arctan}(\sqrt{2}x-1).\end{align}Aaaaaaand here's another

*fun*exercise: differentiate \(\sin(x)\) from first principles! And here's a hint (it's written in white, so highlight it):

You'll probably need to show at some point that \(\lim_{x\to 0}\frac{\sin(x)}{x}=1\)...and this can totes be done by proving the following inequality \(\sin(x)<x<\tan(x)\) for \(x\in(0,\frac{\pi}{2})\). And...this can be done...

*geometrically*.

=P

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