# Question about indefinite integral with square root

I need to integrate the following integral: $$\int{\frac{dx}{x\sqrt{x^2+4x-4}}}.$$ To solve this I used Euler's substitution (i.e. $\sqrt{x^2+4x-4}=x+t$). The result I got is: $\arctan\frac{\sqrt{x^2+4x-4}-x}{2} + C$, this result seems to be correct because after differentiating it I got the original integral. But for some reason the answer in my book is: $\frac{1}{2}\arcsin{\frac{x-2}{x\sqrt{2}}} + C$, which also seems to be correct.

My question is how can I go from the form I got to the other form (which seems to be a lot simpler than the form that I got)? Also, I think I need to use Euler's substition here, because it in chapter that explains it.

• Using trigonometry, evaluate your result. Be sure to know how you treat an arbitrary constant. It should work. – NasuSama Jan 18 '14 at 16:01
• @NasuSama: I tried to use the following identity: $\arctan{x}=\arcsin{\frac{x}{\sqrt{x^2+1}}}$, but without luck. I don't know what else I can do. – EMDB1 Jan 18 '14 at 16:10

The expressions $$F(x) = \tan^{-1} \frac{\sqrt{x^2+4x-4}-x}{2}$$ and $$G(x) = \frac{1}{2} \sin^{-1} \frac{x-2}{x\sqrt{2}}$$ cannot be directly transformed into each other because they are not in fact equal; they differ by a constant of integration on the interval $x \ge 2(\sqrt{2}-1)$. Moreover, on the interval $x < -2(1+\sqrt{2})$, the two functions do not agree: their sum is constant, thus one or the other antiderivative has the incorrect sign. To see which one it is, let's look at $h(x) = \frac{x-2}{x \sqrt{2}}$ on this interval: clearly, it is positive (since if $x < 0$, $h(x) > 0$), and increasing (since $h'(x) > 0$). Furthermore, as $x \to -2(1+\sqrt{2})$ from the left, $h(x) \to 1$. So if we define the inverse sine in the usual way $\sin^{-1} : [-1,1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, then clearly $G(x)$ is an increasing function on $(-\infty, -2(1+\sqrt{2})]$. But this contradicts the requirement that $f(x) = (x \sqrt{x^2+4x-4})^{-1} < 0$ on this interval, for $G(x)$ is increasing implies $G'(x) = f(x) > 0$. Thus the answer in the book is incorrect for $x < -2(1+\sqrt{2})$.

The answer you obtained, $F$, does not have this problem. It is correct for all $x \in (-\infty, -2(1+\sqrt{2})] \cup [2(\sqrt{2}-1), \infty)$.

Now, to show that $F$ and $G$ are equivalent on $[2(\sqrt{2}-1),\infty)$, we claim that $F(x) - G(x) = \frac{\pi}{8}$. Then it suffices to compute $$\tan \left( G(x) + \frac{\pi}{8} \right)$$ and show this is $(\sqrt{x^2+4x-4} - x)/2$, which is an exercise I leave to you.

This discrepancy between $F$ and $G$ that arises out of different methods of integration naturally suggests a question: was there an error in the method that produced $G$? Or was there an assumption that was made that did not actually hold? Is there a simple way to fix $G$ such that it does work on the entire interval for which the integrand is defined? Again, I leave these questions to you as an instructive exercise. I hope your investigation of this particular problem leads you to a deeper understanding of the meaning of antiderivatives, techniques of integration, and how to check if your computations make sense.


Also, \begin{align} t &=\tan\pars{\theta \over 2}={\sin\pars{\theta} \over 2\cos^{2}\pars{\theta/2}} ={\sin\pars{\theta} \over 1 + \cos\pars{\theta}} ={\root{\sec^{2}\pars{\theta} - 1} \over \sec\pars{\theta} + 1} \\[3mm]&={\root{\bracks{\pars{x + 2}/\pars{2\root{2}}}^{2} - 1} \over \pars{x + 2}/\pars{2\root{2}} +1} ={\root{x^{2} + 4x - 4} \over x + 2 + 2\root{2}} \end{align}

$$\color{#00f}{\large% \int{\dd x \over x\root{x^{2} + 4x- 4}} = \arctan\pars{\bracks{\root{2} + 1}\, {\root{x^{2} + 4x - 4} \over x + 2\bracks{\root{2} + 1}}}} + \mbox{a constant}$$

Relations between $\arcsin$ and $\arctan$: