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

Question

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.

Answer 1

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.

Answer 2

Notice

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

Answer 3

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.

Answer 4

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

Answer 5

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

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, I use standard formula $$4∫\frac{dx}{\sqrt{x^2+a^2}} =\ln\left(x+\sqrt{x^2+a^2 }\right)+C, \quad C\in\Bbb R$$

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

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

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

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

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