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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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2 Answers 2

up vote 2 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.

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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 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 at 4:28


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

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completing the square? –  Dan Apr 27 at 12:50
exactly......... –  Henry Lebesgue Apr 27 at 12:53
is my answer still acceptable anyway? –  Dan Apr 27 at 12:56
After completeing the square, substitute $u=x-1/2$. It should be more recognizable then. –  mjh Apr 27 at 12:57

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