I tried to use u-substitution on the following integral, and I don't understand why it doesn't work. $\int\arctan(2x)dx$
I let $u = 2x$, so $du = 2dx$. So $\int\arctan(2x)dx = \frac{1}{2}\int\arctan(u)du$. The antiderivative of arctan is $\frac{1}{x^2 + 1}$, so the integral becomes $\frac{1}{2}*\frac{1}{u^2 + 1} = \frac{1}{8x^2 + 2}$, but this isn't right. Where did I go wrong?
You have calculated wrongly the integral.
Indeed, the derivative of $\operatorname{arctan}(u)$ equals the proposed expression.
In order to solve this integral, it can be done by IBP: \begin{align*} \int\arctan(2x)\mathrm{d}x = x\arctan(2x) - \int\frac{2x}{4x^{2} + 1}\mathrm{d}x \end{align*}
In order to solve the second integral, it may be proceeded as follows: \begin{align*} \int\frac{2x}{4x^{2} + 1}\mathrm{d}x & = \frac{1}{4}\int\frac{8x}{4x^{2} + 1}\mathrm{d}x\\\\ & = \frac{1}{4}\int\frac{\mathrm{d}(4x^{2} +1)}{4x^{2} + 1}\\\\ & = \frac{\ln(4x^{2} + 1)}{4} + C \end{align*}
Gathering the previous results, we achieve the desired expression:
\begin{align*} \int\arctan(2x)\mathrm{d}x = x\arctan(2x) - \frac{\ln(4x^{2} + 1)}{4} + C \end{align*}
Hopefully this helps!