Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
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


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

share|cite|improve this answer
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:

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


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

So the answer you got is also correct.

share|cite|improve this answer

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}

share|cite|improve this answer

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

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


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

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.