Integral of $\frac{1}{\sqrt{x^2-x}}dx$

Question

For a differential equation I have to solve the integral $\frac{dx}{\sqrt{x^2-x}}$. I eventually have to write the solution in the form $ x = ...$ It doesn't matter if I solve the integral myself or if I use a table to find the integral. However, the only helpful integral in an integral table I could find was: $$\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}} \ln \left|2ax + b +2\sqrt{a\left({ax^2+bx+c}\right)}\right|$$ Which would in my case give: $$\frac{dx}{\sqrt{x^2-x}} = \ln \left|2x -1 + 2\sqrt{x^2-x}\right|$$ Which has me struggling with the absolute value signs as I need to extract x from the solution. All I know is that $x<0$ which does not seem to help me either (the square root will only be real if $x<-1$).

Is there some other formula for solving this integral which does not involve absolute value signs or which makes extracting $x$ from the solution somewhat easier? Thanks!

Answer 1

setting $$\sqrt{x^2-x}=t+x$$ we obtain $$x=-\frac{t^2}{2t+1}$$ and $$dx=-\frac{2t(1+t)}{(1+2t)^2}dt$$ and our integral is $$-2\int\frac{1}{2t+1}dt$$

Answer 2

Write

$$ \int \frac{ dx}{\sqrt{x^2 - x }} = \int \frac{ dx}{\sqrt{\left(x - \frac{1}{2}\right)^2 - \left( \frac{1}{2}\right)^2}} = \int \frac{dz}{\sqrt{z^2 - \left( \frac{1}{2}\right)^2}}=\cdots$$

Answer 3

$$ \displaylines{ \sqrt {x^2 - x} = t + x \cr \Leftrightarrow x^2 - x = t^2 + 2xt + x^2 \cr \Leftrightarrow - x = t^2 + 2xt \cr - x = \frac{{t^2 }}{{1 + 2t}} \Rightarrow - dx = \frac{{2\left( {t^2 + 1} \right)}}{{\left( {1 + 2t} \right)^2 }}dt \cr \frac{{dx}}{{\sqrt {x^2 - x} }} = - \frac{{\frac{{2\left( {t^2 + 1} \right)}}{{\left( {1 + 2t} \right)^2 }}dt}}{{t - \frac{{t^2 }}{{1 + 2t}}}} \cr = \frac{{2\left( {t^2 + 1} \right)dt}}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}} \cr \frac{{2\left( {t^2 + 1} \right)}}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}} = \frac{\alpha }{t} + \frac{\beta }{{t + 1}} + \frac{\lambda }{{1 + 2t}} \cr} $$ $$\displaylines{ \frac{{2\left( {t^2 + 1} \right)}}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}} = \frac{\alpha }{t} + \frac{\beta }{{t + 1}} + \frac{\lambda }{{1 + 2t}} \cr = \frac{{\lambda \left( {t^2 + t} \right) + \beta \left( {2t^2 + t} \right) + \alpha \left( {2t^2 + 3t + 1} \right)}}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}} \cr = \frac{{t^2 \left( {\lambda + 2\beta + 2\alpha } \right) + t\left( {\lambda + \beta + 3\alpha } \right) + \alpha }}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}} \cr \Rightarrow \left\{ \begin{array}{l} \alpha = 2 \\ \lambda + \beta + 6 = 0 \\ .\lambda + 2\beta + 4 = 2 \\ \end{array} \right. \Rightarrow \left\{ \begin{array}{l} \alpha = 2 \\ \beta = 4 \\ \lambda = - 10 \\ \end{array} \right. \cr} $$ $$ \int {\frac{{2\left( {t^2 + 1} \right)dt}}{{t\left( {1 + 2t} \right)\left( {t + 1} \right)}}} = 2\ln \left( t \right) + 4\ln \left( {1 + t} \right) - 5\ln \left( {1 + 2t} \right) $$

$$\int {\frac{{dx}}{{\sqrt {x^2 - x} }}} = \ln \left( {\frac{{\left( { - x + \sqrt {x^2 - x} } \right)^2 \left( {1 + \left( { - x + \sqrt {x^2 - x} } \right)} \right)^4 }}{{\left( {1 + 2\left( { - x + \sqrt {x^2 - x} } \right)} \right)^5 }}} \right)$$

Answer 4

$x^2-x\ge0\iff x\not\in(0,1)$. Thus, we need to choose whether $x\ge1$ or $x\le0$.

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For $x\ge1$, let $x=u+\frac12$ and $u=\frac12\sec(\theta)$. We can assume that $\theta\in[0,\frac\pi2)$ since that gives a full range for $x\ge1$. As is shown in this answer, $\int\sec(\theta)\,\mathrm{d}\theta=\log(\sec(\theta)+\tan(\theta))+C$. Therefore, $$ \begin{align} \int\frac{\mathrm{d}x}{\sqrt{x^2-x}} &=\int\frac{\mathrm{d}u}{\sqrt{u^2-\frac14}}\\ &=\int\frac{\sec(\theta)\tan(\theta)\,\mathrm{d}\theta}{\tan(\theta)}\\[9pt] &=\log(\sec(\theta)+\tan(\theta))+C\\[9pt] &=\log\left(2x-1+2\sqrt{x^2-x}\right)+C \end{align} $$ Since $2x-1+2\sqrt{x^2-x}\ge0$ for all $x\ge1$, absolute values are not needed.

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For $x\le0$, we can assume $\theta\in(\frac\pi2,\pi]$. Then, we get $$ \int\frac{\mathrm{d}x}{\sqrt{x^2-x}}=\log\left(1-2x-2\sqrt{x^2-x}\right)+C $$ Since $1-2x-2\sqrt{x^2-x}\ge0$ for all $x\le0$, absolute values are not needed.

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The absolute values are needed only if we want to try to give a solution for all $x\not\in(0,1)$. However, since the interval of integration cannot span $(0,1)$, we must have either $x\ge1$ or $x\le0$.

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Since you have stated that $x\lt0$ in your question, that means we can use the solution for $x\lt0$: $$ \int\frac{\mathrm{d}x}{\sqrt{x^2-x}}=\log\left(1-2x-2\sqrt{x^2-x}\right)+C $$ and we can solve for $x$. If $$ u=\log\left(1-2x-2\sqrt{x^2-x}\right) $$ Then $$ \begin{align} e^u&=1-2x-2\sqrt{x^2-x}\\ \left(e^u+2x-1\right)^2&=4x^2-4x\\ e^{2u}+(4x-2)e^u+1&=0\\ x&=\frac12-\frac{e^u+e^{-u}}4\\ &=\frac12(1-\cosh(u)) \end{align} $$

Answer 5

$\sqrt{x}=z \implies\dfrac{dx}{2\sqrt{x}}=dz$

$\displaystyle\int\dfrac{dx}{\sqrt{x^2-x}}=\displaystyle\int\dfrac{dx}{\sqrt{x(x-1)}}=2\displaystyle\int\dfrac{ dz}{\sqrt{z^2-1}}$

$z=\sec \theta \implies dz=\sec \theta \tan \theta\ d\theta$

$\therefore \displaystyle\int\dfrac{ dz}{\sqrt{z^2-1}}=\displaystyle\int{\sec \theta \ d\theta}=\ln \left\lvert \sec\theta+\tan \theta\right\rvert+C$

$\therefore \displaystyle\int\dfrac{dx}{\sqrt{x^2-x}}=2\ln \left\lvert \sqrt{x}+\sqrt{x-1}\right\rvert+C'$

$$\color{blue}{\forall x \geq 1,\ \ln \left\lvert \sqrt{x}+\sqrt{x-1}\right\rvert=\ln \left( \sqrt{x}+\sqrt{x-1}\right)}$$

Answer 6

$$\int \frac{dx}{\sqrt{x^2-x}} = \int \frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2 - \frac{1}{4}}}$$

Setting $ x - \frac{1}{2} = t $

$$ \int \frac{dx}{\sqrt{\left(x-\frac{1}{2}\right)^2 - \frac{1}{4}}} = \int \frac{dt}{\sqrt{t^2 - \frac{1}{4}}} $$

It's possible to do this using a trig substitution, but if you want the inverse function, a better way is to use a hyperbolic substitution. Let $t = \frac{1}{2}\cosh u \Rightarrow dt = \frac{1}{2} \sinh u$

$$ \int \frac{dt}{\sqrt{t^2 - \frac{1}{4}}} = \int \frac{\frac{1}{2}\sinh u \, du}{\sqrt{\frac{1}{4}(\cosh^2 u - 1)}} = \int \frac{\frac{1}{2}\sinh u \, du}{\frac{1}{2}\sinh u} = \int du = u + C $$

Working backwards is easy since you already have $t$ as a function of $u$ $$t = \frac{1}{2}\cosh u \Rightarrow x = \frac{1}{2}\cosh u + \frac{1}{2}$$

Where $u$ is the integral in question

Answer 7

I'm guessing a bit as to what the OP means, but I think he or she started with the differential equation

$${dx\over dt}=\sqrt{x^2-x}$$

and wishes to find $x$ as a function of $t$. The formalism

$${dx\over\sqrt{x^2-x}}=dt$$

leads to

$$\int{dx\over\sqrt{x^2-x}}=\int dt=t+C$$

so that what the OP cites from the table of integrals says

$$\ln\left|2x-1+2\sqrt{x^2-x}\right|=t+C$$

and this is what the OP wants to solve for $x$.

I'm going to stop at this point until I hear (in a comment) that this is indeed what the OP means.

Answer 8

In general

$$\int \frac{1}{\sqrt{ax^2+bx}}dx = \begin{cases} \frac{-1}{\sqrt{-a}} \tan^{-1} \frac{ax+\frac b2}{\sqrt{-a}\sqrt{ax^2+bx}}, & a<0\\ \frac{1}{\sqrt{a}} \coth^{-1} \frac{ax+\frac b2}{\sqrt{a}\sqrt{ax^2+bx}}, & a>0 \end{cases}$$

Answer 9

$$ I = \int \frac{1}{\sqrt{x^2 -x}}\mathrm dx$$

Let $x=\sec^2\theta$ and we get :

$$I = \int \frac{2\sec\theta(\sec\theta \tan\theta)}{\sec\theta\tan\theta}\mathrm d\theta$$

$$I= 2\int\sec\theta\mathrm d\theta$$

$$I = 2\ln(\sec\theta+\tan\theta) + C$$

Reversing back the substitutions, we get :

$$I = 2\ln(\sqrt{x} + \sqrt{x-1})+C$$