Inverse Tangent of a Rational Function
<p>The given expression is:</p>
<p>\[\tan^{-1}\left(\frac{1+x}{1-x}\right)\]</p>
<p>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 \(\tan(\tan^{-1}(x)) = x\).</p>
<p>\[\tan(y) = \frac{1+x}{1-x}\]</p>
<p>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:</p>
<p>\[h = \sqrt{(1+x)^2 + (1-x)^2}\]</p>
<p>\[h = \sqrt{1+2x+x^2 + 1-2x+x^2}\]</p>
<p>\[h = \sqrt{2x^2 + 2}\]</p>
<p>\[h = \sqrt{2(x^2 + 1)}\]</p>
<p>Now we express \(\sin(y)\) and \(\cos(y)\) in terms of the sides of the triangle:</p>
<p>\[\sin(y) = \frac{opposite}{hypotenuse} = \frac{1+x}{\sqrt{2(x^2 + 1)}}\]</p>
<p>\[\cos(y) = \frac{adjacent}{hypotenuse} = \frac{1-x}{\sqrt{2(x^2 + 1)}}\]</p>
<p>We use the identity \(\tan(y) = \frac{\sin(y)}{\cos(y)}\) to simplify:</p>
<p>\[\tan(y) = \frac{\frac{1+x}{\sqrt{2(x^2 + 1)}}}{\frac{1-x}{\sqrt{2(x^2 + 1)}}}\]</p>
<p>\[\tan(y) = \frac{1+x}{1-x}\]</p>
<p>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.</p>
