# 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.

• 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

## 3 Answers

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.

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}{\left\langle #1 \right\rangle}% \newcommand{\braces}{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}{\displaystyle{#1}}% \newcommand{\equalby}{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}{\left\vert #1\right\rangle}% \newcommand{\ol}{\overline{#1}}% \newcommand{\pars}{\left( #1 \right)}% \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}{\underline{#1}}% \newcommand{\verts}{\left\vert\, #1 \,\right\vert}$ $${\cal I}\equiv\int{\dd x \over x\root{x^{2} + 4x- 4}} =\int{\dd x \over x\root{\pars{x + 2}^{2} - 8}}$$ 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}

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

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