# How to calculate $\int \sqrt{x^2+6}\,dx$?

How to calculate $\int \sqrt{x^2+6}\,dx$, by using Euler substitution and with to use of the formula : $\int u\,dv = vu - \int v\,du$. note: what I mean by Euler substitution: is when we have a Integrand like $\sqrt{x^2+1}$, then we can use the trick of substituting $t= x + \sqrt{x^2+1}$. and that gives us: $dt=\frac{t}{\sqrt{x^2+1}} \, dx$, which can be helpful while solving the questions.

here I supposed that $u=\sqrt{x^2+6}$, and $du= \frac{x}{\sqrt{x^2+6}}\,dx$

then: $\int u\,dv = x\sqrt{x^2+6} - \int \frac{x^2}{\sqrt{x^2+6}}\,dx$, and then I got stuck with the latter integral. trying to substitute $t= x + \sqrt{x^2+6}$ in this case didn't really help. how can I do it $especially$ in this way? I know there are might be a lot of creative solutions, but I want some help continuing that direction.

• I'd think about $\tan^2+\mathbf{1}=\sec^2$. – Git Gud Mar 22 '15 at 16:19
• Hint: $D\,\text{arsinh}\,(x/\sqrt{6})=1/\sqrt{x^2+6}$. (But first you should add and subtract $6$.) In fact, your substitution should work (at least after adding and subtracting $6$). – mickep Mar 22 '15 at 16:23
• Via the given substitution I've reduced it to integrating a simple rational function; see below. ${}\qquad{}$ – Michael Hardy Mar 22 '15 at 17:01

\begin{align} t & = x + \sqrt{x^2+6} \\[8pt] dt & = \left( 1 + \frac x{\sqrt{x^2+6}} \right)\,dx = \frac t {\sqrt{x^2+6}} \,dx \\[8pt] \frac{dt} t & = \frac{dx}{\sqrt{x^2+6}} \end{align} \begin{align} t & = x+\sqrt{x^2+6} \\ 2x-t & = x - \sqrt{x^2+6} \\[6pt] \text{Hence, }t(2x-t) & = -6 \\ 0 & = t^2 - 2xt -6 \\[6pt] x & = \frac{t^2-6}{2t} \\[6pt] \sqrt{x^2+6} & = \frac{t^2+6}{2t} \\[6pt] dx & = \sqrt{x^2+6}\,\frac{dt} t = \frac{(t^2+6)\,dt}{2t^2} \end{align} Therefore $$\int \sqrt{x^2+6}\ dx = \int \frac{t^2+6}{2t} \cdot \frac{(t^2+6)\,dt}{2t^2} \text{ etc.}$$

Let your integral be $I$. Then from what you wrote we have $$I=x\sqrt{x^2+6}-\int\left(\frac{x^2+6}{\sqrt{x^2+6}}-\frac{6}{\sqrt{x^2+6}}\right)\,dx,$$ which we can rewrite as $$I=x\sqrt{x^2+6}-I+6\int \frac{1}{\sqrt{x^2+6}}\,dx.$$ Solve for $I$.

The integral that remains to do is straight Euler substitution $t=x+\sqrt{x^2+6}$. Then $\frac{dx}{\sqrt{x^2+6}}=\frac{dt}{t}$.

• Thats my man! :) Thanks!!! – Firas Ali Abdel Ghani Mar 22 '15 at 16:55
• You are welcome. The approach of Michael Hardy is more directly Euler substitution, so in that sense a better answer to your question. I instead used, along with Euler substitution, your integration by parts. – André Nicolas Mar 22 '15 at 18:05

Forget for a second about stupid number 6. You want a substitution which will make $x^2+1$ a perfect square. Let see. Euler seems overkill for something like that. Recall

$$\cosh^2(x)-\sinh^2(x)=1$$ where $\cosh(x)$ and $\sinh(x)$ stand for hyperbolic function. It should be pretty obvious that you need something like

$$t=\sqrt{6}\sinh(x)$$

• I still didn't learn hyperbolic functions, but for somehow they gave me this questions with the tools I've suggested. how could that be possible? I'm pretty sure there is another way, but thanks for the effort :) I will check your solution thoroughly when I learn about this kind of functions. – Firas Ali Abdel Ghani Mar 22 '15 at 16:38
• It is suffices to use trigonometric substitution. It just doesn't seem natural to my mathematical eye. The question is routine. Your teacher is not asking anything crazy. – Predrag Punosevac Mar 22 '15 at 16:55

For the integral \begin{align} I = \int \sqrt{ x^{2} + 6 } \, dx \end{align} let $x = \sqrt{6} \sinh(t)$ to obtain $dx = \sqrt{6} \cosh(t) dt$ and \begin{align} I &= \int \sqrt{6} \, \sqrt{ 6 ( 1 + \sinh^{2}(t))} \, \cosh(t) \, dt \\ &= 6 \int \cosh^{2}(t) \, dt \\ &= 3 \int ( 1 + \cosh(2t)) \, dt \\ &= 3 \left[ t + \frac{\sinh(2t)}{2} \right] \\ &= 3 \left[ \sinh^{-1}\left( \frac{x}{\sqrt{6}} \right) + \frac{ x \, \sqrt{x^{2} + 6} }{ 6} \right] \end{align}

Notes: Reading the question properly this solution would have been different. With that being said the presentation here is an alternate method of producing a solution to the proposed integral.

• I still didn't learn about hyperbolic functions, but thanks for your effort :) I appreciate it. – Firas Ali Abdel Ghani Mar 22 '15 at 16:39