Integral: $\int \frac{dx}{\sqrt{x^{2}-x+1}}$

How do I integrate this? $$\int \frac{dx}{\sqrt{x^{2}-x+1}}$$ I tried solving it, and I came up with $\ln\left | \frac{2\sqrt{x^{2}-x+1}+2x-1}{\sqrt{3}} \right |+C$. But the answer key says that the answer should be $\sinh^{-1}\left ( \frac{2x-1}{\sqrt{3}} \right )+C$. Any answer is very appreciated.

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Completing the square will yield $$x^2 - x + 1 = \left(x-\frac{1}{2}\right)^2 + \frac{3}{4}$$ Normally, we will let $u=x-\frac{1}{2}$. However it can also be solved by letting $x-\frac{1}{2}=\frac{\sqrt3}{2}\sinh t$ and $dx=\frac{\sqrt3}{2}\cosh t\ dt$ which yields \begin{align} \int \frac{dx}{\sqrt{x^{2}-x+1}}&=\int \frac{\frac{\sqrt3}{2}\cosh t\ dt}{\sqrt{\frac{3}{4}\sinh^2 t+\frac{3}{4}}}\\ &=\int \frac{\cosh t\ dt}{\sqrt{\cosh^2 t}}\\ &=\int \ dt\\ &=t+C \end{align} where $\sinh t=\dfrac{2x-1}{\sqrt3}\;\Rightarrow\; t=\sinh^{-1}\left(\dfrac{2x-1}{\sqrt3}\right)$. Thus $$\int \frac{dx}{\sqrt{x^{2}-x+1}}=\sinh^{-1}\left(\dfrac{2x-1}{\sqrt3}\right)+C.$$ As your book's solution.

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Nice method, I like how you used the hyperbolic substitution and not the linear one you mentioned. +1 – Integrals Jun 4 '14 at 0:49
@Integrals Thanks Jeff. About your comment on the other OP, thanks for your support but I don't wanna involve since if I start argument with other user I had bad experience and often got downvote for no reason. I'll just support you from behind. – Tunk-Fey Jun 4 '14 at 4:28

Notice

$$x^2 - x + 1 = \left(x-\frac{1}{2}\right)^2 + 1 - \frac{1}{4} = \left(x-\frac{1}{2}\right)^2 + \frac{3}{4}$$

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completing the square? – Dan Apr 27 '14 at 12:50
exactly......... – ProbabilityGuy Apr 27 '14 at 12:53
is my answer still acceptable anyway? – Dan Apr 27 '14 at 12:56
After completeing the square, substitute $u=x-1/2$. It should be more recognizable then. – mjh Apr 27 '14 at 12:57
@Dan Your answer is correct. In fact it's the same as the book's. – Dylan Jan 25 '15 at 2:07

Note that these are actually the same answer, since:

$$\sinh^{-1} x = \ln \left(x + \sqrt{x^2+1} \right)$$

and

$$\frac{2x-1}{\sqrt{3}} + \sqrt{\left(\frac{2x-1}{\sqrt{3}}\right)^2 + 1} = \frac{2x - 1 + 2\sqrt{x^2 - x + 1}}{\sqrt{3}}$$

So the answer you got is also correct.

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ The straightforward method was given in @Tunk-Fey answer. However, we just show another method ( one of the Euler sub$\ldots$ ) which will be fine for the OP to know it.

Make the sub $\ds{\root{x^{2} - x + 1} - x \equiv t}$ such that $\ds{x = \frac{1 - t^{2}}{1 + 2t}}$ and \begin{align} \int\frac{\dd x}{\root{x^{2} - x + 1}}&=-\int\frac{2\,\dd t}{2t + 1} =-\ln\pars{2t + 1} \\[5mm]&=-\ln\pars{2\root{x^{2} - x + 1} - 2x + 1} + \mbox{a constant} \end{align}

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We have, $$∫\frac{dx}{\sqrt{x^2-x+1}}=?$$

Now, observe the expression $$x^2-x+1=\left(x^2-x\right)+1=\left(x-\frac{1}{2}\right)^2+1-\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^{2}$$

$$∴ ∫\frac{dx}{\sqrt{x^2-x+1}}=∫\frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2+\left(\frac{3}{2}\right)^2}}$$

Now,use standard formula $∫\frac{dx}{\sqrt{x^2+a^2}} =ln\left(x+\sqrt{x^2+a^2 }\right)+C$

$$∫\frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2+\left(\sqrt{\frac{3}{2}}\right)^2}}=ln\left(\left(x-\frac{1}{2}\right)+\sqrt{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right)$$

$$=ln\left(\frac{2x-1}{2}+\sqrt{\left(\frac{2x-1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right)$$

$$=ln\left(\frac{√3}{2} \left(\frac{2x-1}{\sqrt{3}}+\sqrt{\left(\frac{2x-1}{\sqrt{3}}\right)^2+1}\right)\right)$$
take out $\frac{√3}{2}$

$$=ln\left(\frac{2x-1}{√3}+\sqrt{\left(\frac{2x-1}{\sqrt{3}}\right)^2+1}\right)+ln \frac{\sqrt{3}}{2}$$

$$=sinh^{-1} \left(\frac{2x-1}{√3}\right)+C'$$

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