The integral $\int\frac{2(2y^2+1)}{(y^2+1)^{0.5}} dy$

Question

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.

Answer 1

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.

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(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.)

Answer 2

This can probably be cut down slightly but it's what I've got so far. For the record, I'm not sure why you'd want to avoid trigonometric substitution but it's an interesting challenge anyway.

If we use the substitution $u=\sqrt{y^2+1}$

We have that for our integral:

$$\begin{align} I&=\int{\frac{2(2y^2+1)}{\sqrt{y^2+1}}}\mathrm{d}y \\ &=\pm \int{\frac{2(2u^2-1)}{u}\cdot\frac{u}{\sqrt{u^2-1}}}\mathrm{d}u \\ &=\pm 2\int{\frac{2u^2-1}{\sqrt{u^2-1}}}\mathrm{d}u \\ |I|&= 2\int{\frac{2u^2}{\sqrt{u^2-1}}-\frac{1}{\sqrt{u^2-1}}}\mathrm{d}u \\ &= 2\int{\frac{2u^2}{\sqrt{u^2-1}}\mathrm{d}u-2\int\frac{1}{\sqrt{u^2-1}}}\mathrm{d}u \end{align}$$

We now set $J=\int\frac{1}{\sqrt{u^2-1}}\mathrm{d}u$ and integrate by parts:

$$\begin{align} |I|&=2\int\frac{2u^2}{\sqrt{u^2-1}}\mathrm{d}u-2J \\ &= 4u^2J - 2\int\left(J\cdot\frac{\mathrm{d}}{\mathrm{d}u}(2u^2)\right)\mathrm{d}u-2J \\ &= 2(2u^2-1)J - 8\int Ju\,\mathrm{d}u\end{align}$$

And integrate by parts again:

$$\begin{align} |I|&= 2(2u^2-1)J - 8\left[u\int J \mathrm{d}u - \iint J\,\mathrm{d}u\mathrm{d}u\right] \\ &= 2(2u^2-1)J - 8u\int J \mathrm{d}u +8 \iint J\,\mathrm{d}u\mathrm{d}u\end{align}$$

Now we use Andre Nicolas' answer to How to integrate $\frac{1}{\sqrt{1+x^2}}$ using substitution? to show that $J$ can be integrated with the substitution $x=\frac{1}{2}(t-\frac{1}{t})$ since $\int J\mathrm{d}u=\int\frac{1}{\sqrt{1+x^2}}\mathrm{d}x$ with the substitution $ix=u$, where $i=\sqrt{-1}$.

So we can say that $J=\ln\left(\sqrt{u^2-1}+u\right)+C_1$ and then, again by parts find that:

$$\begin{align} \int J \mathrm{d}u&=C_2 u+C_3-\sqrt{u^2-1}+u \ln(\sqrt{u^2-1}+u) \\ & = C_2u+C_3 - \sqrt{u^2-1}+uJ \end{align} $$

Where all $C_i$ are arbitrary constants.

and: $$\begin{align}\iint J \mathrm{d}u\mathrm{d}u &= C_4 u^2+C_5 u+C_6+\frac{1}{4}\left((2 u^2+1) \ln(\sqrt{u^2-1}+u)-3u\sqrt{u^2-1}\right) \\ &= C_4 u^2+C_5 u+C_6+\frac{1}{4}\left((2 u^2+1) J-3u\sqrt{u^2-1}\right) \end{align}$$

Which gives us a final answer of: $$\begin{align}|I|&=2(2u^2-1)J+C_1 \\ &\quad-8u(C_2 u+C_3-\sqrt{u^2-1}+u J) \\ &\quad+8(C_4 u^2+C_5 u+C_6+\frac{1}{4}\left((2 u^2+1) J-3u\sqrt{u^2-1}\right)) \\ &=2J\left[(2u^2-1)-4u^2+(2u^2+1)\right] \\ &\quad+2(4u-3u)\sqrt{u^2-1} \\ &\quad+2(4C_4-4C_2)u^2 \\ &\quad+2(4C_5-4C_3)u \\ &\quad+2(C_1-4C_6) \\ &=2u\sqrt{u^2-1}+C_7u^2+C_8u+C_9 \end{align}$$

We substitute $y$ back in as $y=\sqrt{u^2-1}$

$$\begin{align}I&=2y\sqrt{1+y^2}+C_7(1+y^2)+C_8\sqrt{1+y^2}+C_9 \\ &=(C_8+2y)\sqrt{1+y^2}+C_7y^2+C_{11}y+C_{10} \end{align}$$

Now if we set all constants as equal to $0$, we get:

$$I=2y\sqrt{1+y^2}$$

So this solution agrees with WolframAlpha and the other answer posted by Alex M. but didn't use any trigonometric or hyperbolic substitutions.

Drop me a comment if you spot an error or steps that can be shortened.

Answer 3

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 $\displaystyle\int \frac{2 (2y^2+1)}{(y^2+1)^{0.5}} dy=2\int\frac{\frac{1}{2}\left(t^2-2+\frac{1}{t^2}\right)+1}{\frac{1}{2}\left(t+\frac{1}{t}\right)}\cdot\frac{1}{2}\left(1+\frac{1}{t^2}\right)dt$

$\displaystyle=2\int\frac{t^2+\frac{1}{t^2}}{t+\frac{1}{t}}\cdot\frac{1}{2}\left(1+\frac{1}{t^2}\right)dt=\int\frac{t^4+1}{t(t^2+1)}\cdot\frac{t^2+1}{t^2}dt$

$\displaystyle=\int\frac{t^4+1}{t^3}dt=\int\left(t+t^{-3}\right)dt=\frac{1}{2}\left(t^2-\frac{1}{t^2}\right)+C=\frac{1}{2}\left(\big(t-\frac{1}{t}\big)\big(t+\frac{1}{t}\big)\right)+C$

$\displaystyle=\frac{1}{2}\left(\big(2y\big)\big(2\sqrt{y^2+1}\big)\right)+C=\color{blue}{2y\sqrt{y^2+1}+C}$

Answer 4

$\displaystyle\int\frac{4y^2+2}{\sqrt{y^2+1}}dy=\int\frac{2y^2+2}{\sqrt{y^2+1}}dy+\int\frac{2y^2}{\sqrt{y^2+1}}dy=2\int\sqrt{y^2+1}dy+\int\frac{2y^2}{\sqrt{y^2+1}}dy.$

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

$\displaystyle2\int\sqrt{y^2+1}dy+2y\sqrt{y^2+1}-2\int\sqrt{y^2+1}dy=\color{red}{2y\sqrt{y^2+1}+C}$