Question - Inverse Tangent of a Rational Function

The given expression is:

\[\tan^{-1}\left(\frac{1+x}{1-x}\right)\]

Let's set \(y = \tan^{-1}\left(\frac{1+x}{1-x}\right)\). To find the value of \(\tan(y)\), we use the fact that \(\tan(\tan^{-1}(x)) = x\).

\[\tan(y) = \frac{1+x}{1-x}\]

We can represent \(y\) as the angle in a right triangle where the opposite side is \(1+x\) and the adjacent side is \(1-x\). To find the hypotenuse \(h\), we use the Pythagorean theorem:

\[h = \sqrt{(1+x)^2 + (1-x)^2}\]

\[h = \sqrt{1+2x+x^2 + 1-2x+x^2}\]

\[h = \sqrt{2x^2 + 2}\]

\[h = \sqrt{2(x^2 + 1)}\]

Now we express \(\sin(y)\) and \(\cos(y)\) in terms of the sides of the triangle:

\[\sin(y) = \frac{opposite}{hypotenuse} = \frac{1+x}{\sqrt{2(x^2 + 1)}}\]

\[\cos(y) = \frac{adjacent}{hypotenuse} = \frac{1-x}{\sqrt{2(x^2 + 1)}}\]

We use the identity \(\tan(y) = \frac{\sin(y)}{\cos(y)}\) to simplify:

\[\tan(y) = \frac{\frac{1+x}{\sqrt{2(x^2 + 1)}}}{\frac{1-x}{\sqrt{2(x^2 + 1)}}}\]

\[\tan(y) = \frac{1+x}{1-x}\]

Which confirms our original expression. To find derivatives or perform integration, we would need additional context, as the problem does not specify an operation to perform with the given expression.