Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|improve this question

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.

share|improve this answer
    
Nice method, I like how you used the hyperbolic substitution and not the linear one you mentioned. +1 –  Integrals Jun 4 at 0:49
1  
@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

Notice

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

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

Your Answer

 
discard

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.