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.
Thank you in advance.
 A: 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.
A: $\newcommand{\+}{^{\dagger}}%
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$$
{\cal I}\equiv\int{\dd x \over x\root{x^{2} + 4x- 4}}
=\int{\dd x \over x\root{\pars{x + 2}^{2} - 8}}
$$
With $x + 2 = 2\root{2}\sec\pars{\theta}$
\begin{align}
{\cal I} &= \int{2\root{2}\sec\pars{\theta}\tan\pars{\theta}
\over \pars{2\root{2}\sec\pars{\theta} - 2}2\root{2}\tan\pars{\theta}}\,\dd\theta=
\half\int{\sec\pars{\theta} \over \root{2}\sec\pars{\theta} - 1}\,\dd\theta
=
\half\int{\,\dd\theta \over \root{2} - \cos\pars{\theta}}
\end{align}
With the Weierstrass substitution $\ds{t = \tan\pars{\theta \over 2}}$:
\begin{align}
{\cal I}&=\half\int{1 \over \root{2} - \pars{1 - t^{2}}/\pars{1 + t^{2}}}
\,{2\,\dd t \over 1 + t^{2}}
=
\int{1 \over \pars{\root{2} + 1}t^{2} + \root{2} - 1}\,\dd t
\\[3mm]&=
{1 \over \root{2} - 1}
\int{1 \over \bracks{\pars{\root{2} + 1}t}^{2} + 1}\,\dd t
=\arctan\pars{\bracks{\root{2} + 1}t}
\end{align}

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}

$$\color{#00f}{\large%
\int{\dd x \over x\root{x^{2} + 4x- 4}}
=
\arctan\pars{\bracks{\root{2} + 1}\,
{\root{x^{2} + 4x - 4} \over x + 2\bracks{\root{2} + 1}}}} + \mbox{a constant}
$$
A: Relations between $\arcsin$ and $\arctan$:
http://www.wolframalpha.com/input/?i=relation+arctan%20x%20%2C+arcsin%20x%20#5016618159685879892.
