How to calculate $\int \sqrt{1+\frac{4x^2}{(1-2x)^2}}dx = $?

Question

I want to find an arc length of y = $\ln{(1-x^2)}$ on the $x\in<0,\frac{1}{2}>$ interval?

$y' = \frac{-2x}{1-x^2}$

So:

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

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

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

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

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

And now I have tried from one of Euler's substitution, which was total nightmare and algebraically impossible for me.

Is there any simpler way to calculate this integral?

Answer 1

In the first equation after "So:", in the numerator, you should have $(1-x^2)^2$ instead of $(1-2x)^2$. Then your function to integrate becomes $$\sqrt{1+\frac{4x^2}{(1-x^2)^2}}=\sqrt{\frac{(1+x^2)^2}{(1-x^2)^2}}$$ You can now take the square root, since $0\lt 1-x^2$ and $0\lt 1+x^2$

Answer 2

Hint:

Use the substitution $$\frac{2x}{1-2x}=\sinh t,\qquad \frac{2\,\mathrm d x}{(1-2x)^2}==\cosh t\,\mathrm dt.$$