Question:

<p><strong>Step 1:</strong> Write tan x as a combination of the denominator and related terms.</p><p>Note that: tan x = (1 + tan x + tan²x) - (1 + tan²x)</p><p>Therefore:</p><p>$$I = \int \frac{(1 + \tan x + \tan^2 x) - (1 + \tan^2 x)}{1 + \tan x + \tan^2 x} dx$$</p><p><strong>Step 2:</strong> Split the integral into two parts.</p><p>$$I = \int 1 \, dx - \int \frac{1 + \tan^2 x}{1 + \tan x + \tan^2 x} dx$$</p><p>$$I = x - \int \frac{\sec^2 x}{1 + \tan x + \tan^2 x} dx$$</p><p><strong>Step 3:</strong> For the second integral, substitute u = tan x, so du = sec²x dx.</p><p>$$I = x - \int \frac{du}{1 + u + u^2}$$</p><p><strong>Step 4:</strong> Complete the square in the denominator.</p><p>$$1 + u + u^2 = \left(u + \frac{1}{2}\right)^2 + \frac{3}{4}$$</p><p><strong>Step 5:</strong> Use the standard form $\int \frac{du}{a^2 + u^2} = \frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right) + C$.</p><p>$$\int \frac{du}{\left(u + \frac{1}{2}\right)^2 + \frac{3}{4}} = \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{u + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2u + 1}{\sqrt{3}}\right)$$</p><p><strong>Step 6:</strong> Substitute back u = tan x.</p><p>$$I = x - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2\tan x + 1}{\sqrt{3}}\right) + C$$</p><p>Comparing with the given form: $I = Kx + A\tan^{-1}\left(\frac{2\tan x + 1}{\sqrt{3}}\right) + C$</p><p>∴ <strong>K = 1 and A = -2/√3 (or -2√3/3)</strong></p>