# What is the integral of dx/(x-2x^2)^(1/2)?

Jun 16, 2015

Let's see if this can be rewritten.

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

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

Completing the Square:
$= \int \frac{1}{\sqrt{- 2 \cdot \left({x}^{2} - \frac{x}{2} + \frac{1}{16} - \frac{1}{16}\right)}} \mathrm{dx}$

$= \int \frac{1}{\sqrt{- 2 \cdot {\left(x - \frac{1}{4}\right)}^{2} + \frac{1}{8}}} \mathrm{dx}$

u-substitution:
Now, let:
$u = x - \frac{1}{4}$
$\mathrm{du} = \mathrm{dx}$

$= \int \frac{1}{\sqrt{\frac{1}{8} - 2 {u}^{2}}} \mathrm{du}$

$= \int \frac{1}{\sqrt{\frac{1}{8} \left(1 - 16 {u}^{2}\right)}} \mathrm{du}$

$= \int \frac{1}{\sqrt{\frac{1}{8}} \sqrt{1 - 16 {u}^{2}}} \mathrm{du}$

$= \int \frac{1}{\sqrt{2} \sqrt{\frac{1}{16} - {u}^{2}}} \mathrm{du}$

$= \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\frac{1}{16} - {u}^{2}}} \mathrm{du}$

Now that it looks better...

Trig substitution:
Let:
$u = a \sin \theta$ with $a = \frac{1}{4}$,
with the form $\sqrt{{a}^{2} - {x}^{2}}$ resembling $\sqrt{1 - {\sin}^{2} \theta}$.
$\mathrm{du} = \frac{1}{4} \cos \theta d \theta$
$\sqrt{\frac{1}{16} - {u}^{2}} = \sqrt{\frac{1}{4} ^ 2 - \frac{1}{4} ^ 2 {\sin}^{2} \theta} = \frac{1}{4} \cos \theta$

We get:

$\frac{1}{\sqrt{2}} \int \frac{1}{\cancel{\frac{1}{4}} \cancel{\cos \theta}} \frac{1}{\cancel{4}} \cancel{\cos \theta} d \theta = \frac{\theta}{\sqrt{2}} + C$

$= \arcsin \frac{4 u}{\sqrt{2}} + C$

but $u = x - \frac{1}{4}$, so $4 u = 4 x - 1$:

$\implies \arcsin \frac{4 x - 1}{\sqrt{2}} + C$

for $x \in \left(0 , \frac{1}{2}\right)$

Since Wolfram Alpha does not give the same answer , I decided to check by taking the derivative of the integration result.

...or:

$\frac{d}{\mathrm{dx}} \left[\arcsin \frac{4 x - 1}{\sqrt{2}} + C\right] = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{1 - {\left(4 x - 1\right)}^{2}}} \cdot 4$

$= \frac{4}{\sqrt{2}} \cdot \frac{1}{\sqrt{1 - 16 {x}^{2} + 8 x - 1}}$

$= {\left(\sqrt{2}\right)}^{3} / \left(\sqrt{- 16 {x}^{2} + 8 x}\right) = \frac{\sqrt{8}}{\sqrt{- 16 {x}^{2} + 8 x}}$

$= \frac{\sqrt{8}}{\sqrt{- 16 {x}^{2} + 8 x}} \cdot \frac{\frac{1}{\sqrt{8}}}{\frac{1}{\sqrt{8}}}$

$= \frac{1}{\frac{1}{\sqrt{8}} \sqrt{- 16 {x}^{2} + 8 x}}$

$= \frac{1}{\sqrt{- \left(\frac{16}{8}\right) {x}^{2} + \left(\frac{8}{8}\right) x}}$

= 1/(sqrt(-2x^2 + x)) = 1/(sqrt(x-2x^2)

again, for $x \in \left(0 , \frac{1}{2}\right)$