Sorry about that. Sometimes posts get lost in the shuffle and days go by before they get noticed. I found an error in your work, in the third step:
\(\displaystyle \displaystyle \frac{2+2x^2-4x^2}{\left(1+x^2\right)^2 \sqrt{\frac{\left(1+x^2\right)^2-\left(2x\right)^2}{\left(1+x^2\right)^2}}}dx = \frac{\left(2-2x^2\right)\left(1+x^2\right)}{\left(1+x^2\right)^2 \sqrt{\left(1+x^2\right)^2-\left(2x\right)^2}}dx\)
I see two errors with that. First, looking just at the numerators, you're saying that:
\(\displaystyle 2+2x^2-4x^2=\left(2-2x^2\right)\left(1+x^2\right)\)
Which is only true for x={-1,0,1}. Now, looking at the denominators:
\(\displaystyle \left(1+x^2\right)^2 \sqrt{\frac{\left(1+x^2\right)^2-\left(2x\right)^2}{\left(1+x^2\right)^2}}=\left(1+x^2\right)^2\sqrt{\left(1+x^2\right)^2-\left(2x\right)^2}\)
Which is also only true for x={-1,0,1}.
If you correct both of these errors, you'd get:
\(\displaystyle \displaystyle \frac{2+2x^2-4x^2}{\left(1+x^2\right)^2 \sqrt{\frac{\left(1+x^2\right)^2-\left(2x\right)^2}{\left(1+x^2\right)^2}}}dx = \frac{2-2x^2}{\left(1+x^2\right)^2 \frac{\sqrt{\left(1+x^2\right)^2-\left(2x\right)^2}}{\sqrt{\left(1+x^2\right)^2}}}dx = \frac{2-2x^2}{\left(1+x^2\right)^2 \frac{\sqrt{\left(1+x^2\right)^2-\left(2x\right)^2}}{\left(1+x^2\right)}}dx = \frac{2-2x^2}{\left(1+x^2\right)^2 \sqrt{\left(1+x^2\right)^2-\left(2x\right)^2}}dx\)
Now try continuing from here and see what you get...
