# Evaluate $\int \frac {\operatorname d\!x} {2x \sqrt{1-x}\sqrt{2-x + \sqrt{1-x}}}$

$$\int \dfrac {\operatorname d\!x} {2x \sqrt{1-x}\sqrt{2-x + \sqrt{1-x}}}$$

Hey there, I've got this complicated integral to evaluate, but I don't know how to go about. I have tried making two substitutions:

1. $t^2 = 1 - x$

2. $x = \sin^2\theta$

But both gave another complicated integral to evaluate:

$$\int \dfrac {\operatorname d\!t} {(t^2-1)\sqrt{ t^2 + t + 1 }}$$

I tried to get the answer for this one using wolfram alpha, but it gave a HUGE, simply HUGE solution. I also tried to get the solution for the original question via wolfram alpha, but it timed out.

Any ideas?

• Where does the integral come from? Do you need the full analytic solution? May 8, 2013 at 14:43
• We had this as a test question. But now I have started doubting that there might be some error in the question, don't know though because no one had announced that there was any error in the question paper. May 8, 2013 at 14:44
• No, hints will do, as they almost always do in my questions :D (don't know what you mean by full analytic solution though!) May 8, 2013 at 14:46
• Mathematica gives for first integral this: $\frac12\left(\frac{\log(1-\sqrt{1-x})}{\sqrt3}-\log(1+\sqrt{1-x})+\log(1-\sqrt{1-x}+2\sqrt{2+\sqrt{1-x}-x})-\frac{\log(3+3\sqrt{1-x}+2\sqrt3\sqrt{2+\sqrt{1-x}-x})}{\sqrt3}\right)$ May 8, 2013 at 14:49
• You are getting a log. Means : USE partial fractions. Seems absurd, though. May 8, 2013 at 14:53

For what it's worth, you can break things up a bit by recognizing that

$$\frac{1}{t^2-1} = \frac12 \left ( \frac{1}{t-1} - \frac{1}{t+1}\right )$$

For example, consider the $t-1$ piece; you may substitute $u=t+1/2$ and get

$$\int \frac{dt}{(t-1) \sqrt{t^2+t+1}} = \int \frac{du}{(u-3/2) \sqrt{u^2+3/4}}$$

This latter integral is relatively tame according to WA:

$$\frac{1}{\sqrt{3}} \left[\log{\left(u-\frac{3}{2}\right)}-\log{\left(\sqrt{12 u^2+9}+3 u+\frac{3}{2}\right)}\right]+C$$

where $C$ is a constant of integration. A similar expression may be found for the $t+1$ piece.

• I think that's a good way to go...! May 8, 2013 at 15:41

The following integral

$$\int \frac{1}{x \sqrt{ax^2 + bx + c}}\text{d}x$$

can be solved by substituting $x = \frac{1}{t}$ to get

$$\int \frac{\pm 1}{\sqrt{a + bt + ct^2}}\text{d}t$$

Which can be recast as the derivative of an inverse trigonometric function (could be hyperbolic, depending on the signs taken):

$$\int \frac{\pm 1}{\sqrt{\pm 1 \pm x^2}}$$

And as noted by Ron, your integral is the sum of two such integrals.

The $$t$$-integral is primed for the well-known Euler substitution. Let

$$u=\frac{\sqrt{t^2+t+1} - 1}t \implies t = -\frac{1-2u}{1-u^2} \implies dt = \frac{2u^2-2u+2}{(1-u^2)^2} \, du$$

to transform to

\begin{align*} I &= \int \dfrac{dt}{(t^2-1)\sqrt{ t^2 + t + 1 }} \\[1ex] &= 2 \int \frac{u^2 - u + 1}{(1-u^2)^2 \left(\frac{(1-2u)^2}{(1-u^2)^2} - 1\right) \left(1-\frac{u-2u^2}{1-u^2}\right)} \, du \\[1ex] &= 2 \int \frac{u^2-1}{u(u-2)(u^2+2u-2)} \, du \end{align*}

Apply the Euler substitution $$\sqrt{t^2+t+1}=(t+\frac12)y+\frac{\sqrt3}2$$.

\begin{align} & \int \frac1{(t^2-1)\sqrt{ t^2 + t + 1 }} \ dt\\ =&\int \frac{2}{3y^2+2\sqrt3 y-3} + \frac{2}{y^2-2\sqrt3 y-1} \ dy\\ =& -\frac1{\sqrt3}\tanh^{-1}\frac{\sqrt3 y+1}{2}-\tanh^{-1}\frac{y-\sqrt3}{2}\\ =&\ -\frac1{\sqrt3}\tanh^{-1}\frac{\sqrt3 \sqrt{t^2+t+1}+t-1}{2t+1} - \tanh^{-1}\frac{\sqrt{t^2+t+1}-\sqrt3(t+1)}{2t+1} \end{align}