The integral $\int\frac{2(2y^2+1)}{(y^2+1)^{0.5}} dy$ What is $$\int\frac{2(2y^2+1)}{(y^2+1)^{0.5}} dy?$$ I split it as $\frac{y^{2}}{(y^2+1)^{0.5}} + \sqrt{y^2+1}.$ Now I substituted $y^{2}=u $ thus $2y\,dy=du$ so we get $0.5 \sqrt{\frac{u}{u + 1}} + 0.5 \sqrt{\frac{1 + u}{u}}$ but now what to do? Another idea was doing $+1-1$ in original question but that too doesn't lead anywhere. Now $y=\tan{x} $ as suggested below is an easy way but I am seeking for a purely algebraic way. Thanks.

$\displaystyle\int\frac{4y^2+2}{\sqrt{y^2+1}}dy=\int\frac{2y^2+2}{\sqrt{y^2+1}}dy+\int\frac{2y^2}{\sqrt{y^2+1}}dy=2\int\sqrt{y^2+1}dy+\int\frac{2y^2}{\sqrt{y^2+1}}dy.$
Now use $\displaystyle u=2y,\;dv=\frac{y}{\sqrt{y^2+1}}dy\;$ so $\;du=2dy,\;v=\sqrt{y^2+1}$ in the 2nd integral to obtain
$\displaystyle2\int\sqrt{y^2+1}dy+2y\sqrt{y^2+1}-2\int\sqrt{y^2+1}dy=\color{red}{2y\sqrt{y^2+1}+C}$
