부분적분법

We integrate by parts, using $u = \arctan(2y)$, $\mathrm{d}v = 4y \,\mathrm{d}y$, so that $v=2y^2$ and $\mathrm{d}u= \frac{2\, \mathrm{d}y}{1+(2y)^2}$:

$$\begin{aligned} \int 4y\arctan(2y) \,\mathrm{d}y = \underbrace{2y^2\arctan(2y)}_{uv} - \int\underbrace{ \frac{4y^2}{(2y)^2+1} \,\mathrm{d}y}_{v\mathrm{d}u} \end{aligned}$$

The integrand $\frac{4y^2}{(2y)^2+1}$ is a rational function. So the remaining integral can be evaluated using the method of partial fractions, starting with long division. But it is easier to just notice that $\frac{4y^2}{4y^2+1} = \frac{4y^2+1}{4y^2+1} - \frac{1}{4y^2+1}$. We therefore have:

$$\begin{aligned} \int \frac{4y^2}{4y^2+1} \,\mathrm{d}y = \int \left(1 - \frac{1}{4y^2+1}\right)\,\mathrm{d}y = y - \frac12\arctan(2y) + C \end{aligned}$$

The final answer is then

$$\begin{aligned} \int 4y\arctan(2y) \,\mathrm{d}y = 2y^2\arctan(2y) - y + \frac{1}{2}\arctan(2y) + C \end{aligned}$$