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$\int \dfrac {dx}{\sqrt{4x^{2}+1}}$

I've been up to this one for quite a while already, and have tried several ways to integrate it, using substituion, with trigonometric as well as hyperbolic functions. I know(I think) I'm supposed to obtain:

$\dfrac {1}{2}\ln \left| 2\sqrt {x^2 +\dfrac {1}{4}}+2x\right|+c$

However, whatever idea I come up with to try to solve it, it seems I always obtain:

$\dfrac {1}{2}\ln \left| x + \sqrt {x^2 + 1} \right| + c$

Thanks for reading, and hopefully answering too.

Cheers

Yann

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    $\begingroup$ An alternative approach would be $2x=\sinh t$ $\endgroup$ – Lucian Sep 27 '14 at 16:45
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Hint: Use trigonometric substitution and put $$2x = \tan\theta^{(\dagger)} \implies 2\,dx = \sec^2\theta\,d\theta \iff dx = \frac 12 \sec^2 \theta\,d\theta$$

$(\dagger)\;\;\theta = \arctan(2x)$.

Recall, also, that $\tan^2 \theta + 1 = \sec^2\theta$.

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Let $x = \frac{1}{2} \tan\theta$ so that $dx=\frac{1}{2}\sec^2\theta d\theta$. Then

$\begin{align*} \int\frac{dx}{\sqrt{4x^2+1}} &= \frac{1}{2} \int \frac{\sec^2 \theta}{\sec\theta} d\theta \\ &= \frac{1}{2} \int\sec\theta d\theta \\ &= \frac{1}{2} \log|\sec\theta + \tan\theta| + C \end{align*}$

for some real constant $C$. Note that we assumed $\tan\theta = 2x$, so $\sec\theta = \sqrt{4x^2 - 1}$. Hence,

$$\int\frac{dx}{\sqrt{4x^2+1}}= \frac{1}{2}\log|\sqrt{4x^2-1} + 2x| + C.$$

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    $\begingroup$ I appreciate your work but please use $\LaTeX$ to answer the questions here. Here is the link where you can learn $\LaTeX$ :meta.math.stackexchange.com/q/5020/133248 $\endgroup$ – Anastasiya-Romanova 秀 Sep 27 '14 at 16:13
  • $\begingroup$ I reformatted your (excellent) answer using $\LaTeX$. Of course, it's your post so modify it as you see fit. $\endgroup$ – Gamma Function Sep 27 '14 at 17:19
  • $\begingroup$ @GammaFunction thanks for edit and sorry as I just start using this website recently so I am still not very familiar with LATEX $\endgroup$ – Alan Wang Sep 27 '14 at 17:32
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Since it hasn't been mentioned yet, one more way to do this integral is via Euler substitutions. Compared to the other two methods already described, this method is probably the most direct means of obtaining the particular form of the anti-derivative mentioned in the OP.

Using an Euler substitution of the first kind, $\sqrt{4x^2+1}=-2x+t$, we have $x=\frac{t^2-1}{4t}$ and $\mathrm{d}x=\frac{1+t^2}{4t^2}\,\mathrm{d}t$, and hence:

$$\begin{align} \int\frac{\mathrm{d}x}{\sqrt{4x^2+1}} &=\int\frac{2t}{1+t^2}\cdot\frac{1+t^2}{4t^2}\,\mathrm{d}t\\ &=\int\frac{\mathrm{d}t}{2t}\\ &=\frac12\ln{(t)}+\color{grey}{constant}\\ &=\frac12\ln{\left(\sqrt{4x^2+1}+2x\right)}+\color{grey}{constant}. \end{align}$$

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we have $\int \dfrac {dx}{\sqrt{4x^{2}+1}}=\int \dfrac {x}{x\sqrt{4x^{2}+1}}\,dx$.Now let $1+4x^{2}=u^2$, so we have $xdx=\frac{1}{4}udu$ and so $$\int \dfrac {x}{x\sqrt{4x^{2}+1}}\,dx=\frac{1}{4}\int \dfrac {u}{u\frac{u^2-1}{4}}\,du=\int \dfrac {1}{u^2-1}\,du$$ also $$\int \dfrac {1}{u^2-1}\,du=\frac{1}{2}\int (\frac{1}{u-1}-\frac{1}{u+1})\,du=\frac{1}{2}\ln|\frac{u-1}{u+1}|.$$ Therefore $$\int \dfrac {dx}{\sqrt{4x^{2}+1}}=\frac{1}{2}\int (\frac{1}{u-1}-\frac{1}{u+1})\,du=\frac{1}{2}\ln|\frac{u-1}{u+1}|+C.$$

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