## Calculus: Early Transcendentals 8th Edition

$$\int\tan^{-1}2y dy=y\tan^{-1}(2y)-\frac{1}{4}\ln(1+4y^2)+C$$
$$A=\int\tan^{-1}2y dy$$ We would choose $u=\tan^{-1}2y$ and $dv=dy$ For $u=\tan^{-1}(2y)$, according to Chain Rule, we have $$du=\frac{d[\tan^{-1}(2y)]}{dy}dy=\frac{d[\tan^{-1}(2y)]}{d(2y)}\frac{d(2y)}{dy}dy=[\frac{1}{1+(2y)^2}\times2]dy=\frac{2}{1+4y^2}dy$$ For $dv=dy$, then $v=y$ Apply Integration by Parts to A, we have $$A=uvâˆ’\int vdu$$ $$A=y\tan^{-1}(2y)-\int\frac{2y}{1+4y^2}dy$$ $$A=y\tan^{-1}(2y)-2\int\frac{y}{1+4y^2}dy$$ Now is the time for Substitution Rule. Let $z=1+4y^2$. Then $dz=8ydy$. That makes $ydy=\frac{1}{8}dz$ Therefore, $$A=y\tan^{-1}(2y)-2\times\frac{1}{8}\int\frac{1}{z}dz$$ $$A=y\tan^{-1}(2y)-\frac{1}{4}\ln|z|+C$$ $$A=y\tan^{-1}(2y)-\frac{1}{4}\ln|1+4y^2|+C$$ However, as we already know, for all $y\in R$, $y^2\ge0$, or that means $1+4y^2\gt0$. So, for all $y\in R$, in other words, $|1+4y^2|=1+4y^2$ Therefore, $$A=y\tan^{-1}(2y)-\frac{1}{4}\ln(1+4y^2)+C$$