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.
 A: Let us use the substitution $y = \sinh t$ (the hyperbolic sine function). Then $\Bbb d y = \cosh t \ \Bbb d t$, so taking into account the hyperbolic trigonometric identities $\cosh^2 t- \sinh^2 t = 1$ and $2 \sinh^2 t = \cosh 2t - 1$, and the fact that $\cosh t > 0 \ \forall t$, the integral becomes
$$2 \int \frac {2 \sinh^2 t + 1} {\cosh t} \cosh t \ \Bbb d t = 2 \int \cosh 2t \ \Bbb d t = \sinh 2t + C = 2 \sinh t \cosh t + C.$$
Since $t = \text{arcsinh } y$, it remains to come up with a nice result for $\cosh \text{arcsinh } y$. But
$$\cosh \text{arcsinh } y = \sqrt{1 + \sinh^2 \text{arcsinh } y} = \sqrt {1 + y^2} ,$$
so your integral is $2 y \sqrt {1 + y^2} + C$, where $C$ is an arbitrary integration constant.

(Now if anybody ever asks you what hyperbolic trigonometry is good for, you may answer that it allows you to quickly compute integrals, at the very least.)
A: Let $\displaystyle y=\frac{1}{2}\left(t-\frac{1}{t}\right),\;dy=\frac{1}{2}\left(1+\frac{1}{t^2}\right)dt$, 
so that $\displaystyle\sqrt{y^2+1}=\frac{1}{2}\left(t+\frac{1}{t}\right),\;\;t=y+\sqrt{y^2+1},\;\;\frac{1}{t}=\sqrt{y^2+1}-y.$
Then $\displaystyle\int \frac{2 (2y^2+1)}{(y^2+1)^{0.5}} dy=2\int\frac{\frac{1}{2}\left(t^2-2+\frac{1}{t^2}\right)+1}{\frac{1}{2}\left(t+\frac{1}{t}\right)}\cdot\frac{1}{2}\left(1+\frac{1}{t^2}\right)dt$
$\displaystyle=2\int\frac{t^2+\frac{1}{t^2}}{t+\frac{1}{t}}\cdot\frac{1}{2}\left(1+\frac{1}{t^2}\right)dt=\int\frac{t^4+1}{t(t^2+1)}\cdot\frac{t^2+1}{t^2}dt$
$\displaystyle=\int\frac{t^4+1}{t^3}dt=\int\left(t+t^{-3}\right)dt=\frac{1}{2}\left(t^2-\frac{1}{t^2}\right)+C=\frac{1}{2}\left(\big(t-\frac{1}{t}\big)\big(t+\frac{1}{t}\big)\right)+C$
$\displaystyle=\frac{1}{2}\left(\big(2y\big)\big(2\sqrt{y^2+1}\big)\right)+C=\color{blue}{2y\sqrt{y^2+1}+C}$
A: $\displaystyle\int\frac{4y^2+2}{\sqrt{y^2+1}}dy=\int\frac{2y^2+2}{\sqrt{y^2+1}}dy+\int\frac{2y^2}{\sqrt{y^2+1}}dy=2\int\sqrt{y^2+1}dy+\int\frac{2y^2}{\sqrt{y^2+1}}dy.$
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
$\displaystyle2\int\sqrt{y^2+1}dy+2y\sqrt{y^2+1}-2\int\sqrt{y^2+1}dy=\color{red}{2y\sqrt{y^2+1}+C}$
