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

Question

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.

Answer 1

\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 $$ \int \sqrt{x^2+6}\ dx = \int \frac{t^2+6}{2t} \cdot \frac{(t^2+6)\,dt}{2t^2} \text{ etc.} $$

Answer 2

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.

Answer 3

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

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

\begin{equation} t=\sqrt{6}\sinh(x) \end{equation}

Answer 4

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 $\frac{dx}{\sqrt{x^2+6}}=\frac{dt}{t}$.