$~\frac{\mathrm{d}}{\mathrm{dy}}\operatorname{arctg}\left(y\right)~$with integration where$~y~$is a function of$~x~$ $$A:=\int_{}^{}{1\over 4x^2+4x+2}\,\mathrm{dx}= \underbrace{{1\over 2}\operatorname{arctg}\left(2x+1\right)+\text{const}}_{\text{I want to derive this.} }  \tag{1}$$
$$4x^2+4x+2=(2x+1)^2+1\tag{2}$$
$$\therefore~~\int_{}^{}{1\over 4x^2+4x+2}\,\mathrm{dx}=\int_{}^{}{1\over(2x+1)^2+1}\,\mathrm{dx}\tag{3}$$
$$y:=(2x+1)\tag{4}$$
$$A=\int_{}^{}{1\over 1+y^2}\,\mathrm{dx}\tag{5}$$
$$\underbrace{\frac{\mathrm{d}}{\mathrm{dx}}\operatorname{arctg}\left(x\right)={1\over 1+x^2}}_{\text{general formula}}\tag{6}$$
$$\therefore~~A=\int_{}^{}{\mathrm{d}\over\mathrm{dy}}\left(\operatorname{arctg}\left(y\right)\right)\,\mathrm{dx}\tag{7}$$
$$\int_{}^{}{\mathrm{d}\over\mathrm{dy}}\left(\operatorname{arctg}\left(y\right)\right)\,\mathrm{dx}=\underbrace{{1\over 2}\operatorname{arctg}\left(y\right)+\text{const}}_{\text{How this can be derived?}}\tag{8}$$
The official book didn't notate$~y=2x+1~$by the way.
 A: $$\int_{}^{}{\mathrm{d}\over\mathrm{dy}}\left(\operatorname{arctg}\left(y\right)\right)\,\mathrm{dx}\tag{1}$$
$$y=2x+1\tag{2}$$
$$\frac{\mathrm dy}{\mathrm dx}=2~~\Leftrightarrow~~\mathrm{dy}=2\mathrm{dx}~~\Leftrightarrow~~{\mathrm{dy}\over 2}=\mathrm{dx}\tag{3}$$
$$\therefore~~\int_{}^{}{\mathrm{d}\over\mathrm{dy}}\left(\operatorname{arctg}\left(y\right)\right)\,\mathrm{dx}=\int_{}^{}\left({\mathrm{d}\over\mathrm{dy}}\left(\operatorname{arctg}\left(y\right)\right)\right)\,{\mathrm{dy}\over 2}\tag{4}$$
$$={1\over 2}\int_{}^{}\left({\mathrm{d}\over\mathrm{dy}}\left(\operatorname{arctg}\left(y\right)\right)\right)\,\mathrm{dy}\tag{5}$$
$$={1\over 2}\operatorname{arctg}\left(y\right)+\text{const}\tag{6}$$
$$={1\over 2}\operatorname{arctg}\left(2x+1\right)+\text{const}\tag{7}$$
A: 
Let $2x+1=\tan\theta$. Then $\dfrac{1}{4x^2+4x+1}=\dfrac{1}{\sec^2\theta}$ and $2x\,dx=\sec^2\theta\,d\theta$
Therefore
\begin{eqnarray}
\int\dfrac{1}{4x^2+4x+2}\,dx&=&\frac{1}{2}\int\frac{1}{\sec^2\theta}\cdot\sec^2\theta\,d\theta\\
&=&\frac{1}{2}\theta+C\\
&=&\frac{1}{2}\arctan(2x+1)+C
\end{eqnarray}
