<s>[INST] Taylor/Maclaurin series of $\arctan(e^x - 1)$ Right, so this keeps bugging me, and I'm probably stuck in some tunnel trying the same thing over and over again.
Give the Maclaurin series of the function $\arctan(e^x - 1)$. up to terms of degree three. Since I try to be as lazy as possible and differentiating this thing sounds like a lot of work, I want to cheat!
Knowing that the Maclaurin series for $\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} + ....$ and the Maclaurin series for $e^x - 1 = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + ..... $
So yeah, substition to the resque, $\arctan(e^x - 1)$, use the known series and substite in the known series for $e^x - 1$:
$$[1 + x + \frac{1}{2}x^2 + \frac{1}{3!}x^3] - \frac{[1 + x]}{3}^3 + \frac{[1]}{5!}^5$$
Something along the lines of:
$$[1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3] - (\frac{1 + 3x + 3x^2 + x^3}{3})$$
Okay, lets all get 6's:
$$[\frac{6}{6} + \frac{x}{6} + \frac{3}{6}x^2 + \frac{1}{6}x^3] - (\frac{2}{6} + \frac{18x}{6} + \frac{18x^2}{6} + \frac{2x^3}{6})$$
Great... Lets do some algebruh
$$\frac{6}{6} + \frac{x}{6} + \frac{3}{6}x^2 + \frac{1}{6}x^3  - \frac{18x}{6} - \frac{18x^2}{6} - + \frac{1}{6}x^3$$
Join some like terms
$$\frac{6}{6} - \frac{2}{6} + \frac{x}{6} - \frac{18x}{6} + \frac{3}{6}x^2 -  \frac{18x^2}{6} + \frac{1}{6}x^3 - \frac{2}{6}x^3$$
This is so far removed from any of the possible answers and I've spent lots of time on this  one, so it's time to get some real help.
Please help me :)
 [/INST] The correct expansion should be $e^x-1=\sum_{i=1}^\infty \frac{x^i}{i!}$.
Hence substituting it, we get
\begin{align}
\arctan (e^x - 1) &\approx (x+\frac{x^2}{2}+\frac{x^3}{6}) - \frac{(x+\frac{x^2}{2}+\frac{x^3}{6})^3}{3} \\
&\approx (x+\frac{x^2}{2}+\frac{x^3}{6}) - \frac{x^3}{3}\\
&= x+\frac{x^2}2 - \frac{x^3}6
\end{align}
</s>