Integration of expression with root of second order polynomial in denominator

**ironband** · February 22nd 2011, 12:41 PM

Hi all,

I'm trying to help a friend out and a bit rusty on my calc. Trying to find indefinite integral:

$\int \frac{1}{(x-1)\sqrt{4x^2-8x+3}}dx$

I've run it through Wolfram Alpha, but the solution seems overly complex to me...it seems like I should be able to get there a little easier than that, but I could be wrong.

My out of practice intuition says that there might be some way to solve this thing by doing a u substitution for $4x^2-8x+3$ , such that $\frac{du}{dx}=8x-8=8(x-1)$ , and relating that back somehow to the x-1 term...but for the life of me I'm just not getting there.

Anyway, if I really need to go the route of completing the square and making a second substitution as Wolfram suggests, that's fine...but I can't shake the feeling that a more elegant solution exists.

Thanks in advance for any help!

Andre

**ironband** · February 22nd 2011, 01:25 PM

OK...I think it is coming to me...but not sure...

If I complete the square of $4x^2-8x+3$ by subtracting 1, I get

$\int \frac{1}{(x-1)\sqrt{(2x-2)^2-1}}dx$

Then, if I pull a factor of 4 out of the expression under the root

$\sqrt{(2x-2)^2-1}=\sqrt{4(x-1)^2-4\frac{1}{4}}=\sqrt{4[(x-1)^2-\frac{1}{4}]}=2\sqrt{(x-1)^2-\frac{1}{4}}$

Which allows me to write this thing as

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

and if I set u=x-1 and a=1/2, I can use this from my table of integrals:

$\int \frac{1}{u\sqrt{u^2-a^2}}du=\frac{1}{a}arcsec\frac{u}{a}+C$

Does that work, or did I project some wishful thinking into this?

**Soroban** · February 22nd 2011, 01:28 PM

Hello, ironband!

Well done! . . . Great work!

Here's how I reasoned it out . . .

$\int \dfrac{dx}{(x-1)\sqrt{4x^2-8x+3}}$

Under the radical, we have:

. . $4x^2 - 8x + 3 \;=\;4x^2-8x + 4 - 1$

. . . . . . . . . . . . $=\;4(x^2 - 2x + 1) - 1$

. . . . . . . . . . . . $=\;[2(x-1)]^2-1$

$\displaystyle \text{The integral becoms: }\;\int\frac{dx}{(x-1)\sqrt{[2(x-1)]^2-1}}$

$\text{Let: }\:2(x-1) \:=\:\sec\theta \quad\Rightarrow\quad x-1 \:=\:\tfrac{1}{2}\sec\theta$

. . $\Rightarrow\quad dx \:=\:\frac{1}{2}\sec\theta \tan \theta\, d\theta \quad\Rightarrow\quad \sqrt{2(x-1)^2-1} \:=\:\tan\theta$

$\displaystyle \text{Substitute: }\;\int\frac{\frac{1}{2}\sec\theta\tan\theta\, d\theta}{\frac{1}{2}\sec\theta\,\tan\theta} \;=\;\int d\theta \;=\;\theta + C$

$\text{Back-substitute: }\:\text{arcsec }2(x-1) + C$

**ironband** · February 22nd 2011, 01:47 PM

OK, I get the same answer. good deal! Thanks

**TheCoffeeMachine** · February 22nd 2011, 02:04 PM

Note that $(4x^2-8x+3)' = 8(x-1)$ , so let:

$t = \sqrt{4x^2-8x+3} \Rightarrow \begin{cases}<br /> \frac{dt}{dx} = \frac{8(x-1)}{2\sqrt{4x^2-8x+3}} \Rightarrow dx = \frac{\sqrt{4x^2-8x+3}}{4(x-1)}\;{dt} \\<br /> t^2 = 4(x-1)^2-1 \Rightarrow (x-1)^2 = \frac{1}{4}(t^2+1).<br /> \end{cases}$ , then:

$\displaystyle \begin{aligned} I & = \frac{1}{4}\int\frac{\sqrt{4x^2-8x+3}}{(x-1)^2\sqrt{4x^2-8x+3}}\;{dt} = \frac{1}{4}\int\frac{4}{t^2+1}\;{dt} \\& = \int\frac{1}{t^2+1}\;{dt} = \arctan{t} = \arctan{\sqrt{4x^2-8x+3}}+k. \end{aligned}$

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