What is the answer to the sequence e^ (ln (1 - x ^ 2) - ln (x + 1) = 2?

 Absolutely! We should go through the arrangement bit by bit.

Given condition:

\[e^{\ln(1 - x^2) - \ln(x + 1)} = 2\]

Stage 1: Consolidate logarithms utilizing the properties of logarithms:

\[\ln\left(\frac{1 - x^2}{x + 1}\right) = 2\]

Stage 2: Exponentiate the two sides to take out the logarithm:

\[e^{\ln\left(\frac{1 - x^2}{x + 1}\right)} = e^2\]

This rearranges to:

\[\frac{1 - x^2}{x + 1} = e^2\]

Stage 3: Increase the two sides by \(x + 1\) to clear the portion:

\[1 - x^2 = e^2 \cdot (x + 1)\]

Stage 4: Extend and revise the terms:

\[1 - x^2 = e^2x + e^2 - e^2\]

Stage 5: Set the condition to nothing:

\[0 = e^2x - x^2 + e^2 - 1\]

Stage 6: Rework to a quadratic structure:

\[x^2 - e^2x + (1 - e^2) = 0\]

Stage 7: Tackle the quadratic condition. You can utilize the quadratic recipe:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

For our situation, where \(a = 1\), \(b = - e^2\), and \(c = (1 - e^2)\), the arrangements are:

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

Stage 8: Rearrange and assess the articulation for \(x\).