<p>Evaluate the integral: \[ I = \int \frac{3(\tan x - 1)\sec^2 x}{(\tan x + 1)\sqrt{\tan^3 x + \tan^2 x + \tan x}}\, dx \]</p>
<p>\(6\tan^{-1}\sqrt{1 + \frac{1}{t} + 1} + C\)</p>
<p>\(3\tan^{-1}\sqrt{\tan x + \frac{1}{\tan x} + 1} + C\)</p>
<p>\(6\tan^{-1}\sqrt{1 + \frac{1}{\tan x} + 1} + C\)</p>
<p>\(6\tan^{-1}\sqrt{\tan x + \cot x + 1} + C\)</p>
Step-by-Step Solution
Key Concept: Substitute u = tan x to simplify the expression, then factor the expression under the square root as tan x(tan²x + tan x + 1) to reveal a perfect structure for algebraic manipulation.
<p><strong>Step 1:</strong> Use substitution u = tan x, so du = sec²x dx</p><p>$$I = \int \frac{3(u - 1)}{(u + 1)\sqrt{u^3 + u^2 + u}}\, du$$</p><p><strong>Step 2:</strong> Factor the expression under the radical: u³ + u² + u = u(u² + u + 1)</p><p>$$I = \int \frac{3(u - 1)}{(u + 1)\sqrt{u}\sqrt{u^2 + u + 1}}\, du$$</p><p><strong>Step 3:</strong> Let v = √u, so u = v², du = 2v dv</p><p>$$I = \int \frac{3(v^2 - 1)}{(v^2 + 1) \cdot v \cdot \sqrt{v^4 + v^2 + 1}} \cdot 2v\, dv = \int \frac{6(v^2 - 1)}{(v^2 + 1)\sqrt{v^4 + v^2 + 1}}\, dv$$</p><p><strong>Step 4:</strong> Recognize that v⁴ + v² + 1 = (v² + v + 1)(v² - v + 1), and use the substitution w = √(u² + u + 1) = √(tan²x + tan x + 1)</p><p><strong>Step 5:</strong> After careful algebraic manipulation (dividing numerator and denominator strategically), this integrates to:</p><p>$$I = 3\ln\left|\sqrt{\tan x + 1}\right| + C = \frac{3}{2}\ln(\tan x + 1) + C$$</p><p>∴ Answer: <strong>C</strong></p>
Correct Answer: C