Key Concepts and Formulas

• Trigonometric Substitution: Using trigonometric identities to simplify integrals, particularly $\sec^2 x = 1 + \tan^2 x$.
• Partial Fraction Decomposition: Breaking down a rational function into simpler fractions to make integration easier.
• Standard Integral: $\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$.

Step-by-Step Solution

Step 1: Substitution We are given the integral: $\int \frac{\tan x}{1 + \tan x + \tan^2 x} dx$ Let $t = \tan x$. Then, $\frac{dt}{dx} = \sec^2 x = 1 + \tan^2 x = 1 + t^2$. Therefore, $dx = \frac{dt}{1 + t^2}$. Substituting these into the integral, we get: $\int \frac{t}{1 + t + t^2} \cdot \frac{1}{1 + t^2} dt = \int \frac{t}{(1 + t + t^2)(1 + t^2)} dt$ This substitution allows us to work with a rational function in terms of $t$ rather than a trigonometric function.

Step 2: Partial Fraction Decomposition Now we decompose the integrand into partial fractions. We seek constants A, B, C, and D such that: $\frac{t}{(1 + t + t^2)(1 + t^2)} = \frac{At + B}{1 + t + t^2} + \frac{Ct + D}{1 + t^2}$ Multiplying both sides by $(1 + t + t^2)(1 + t^2)$, we have: $t = (At + B)(1 + t^2) + (Ct + D)(1 + t + t^2)$ $t = At + At^3 + B + Bt^2 + Ct + Ct^2 + Ct^3 + D + Dt + Dt^2$ $t = (A + C)t^3 + (B + C + D)t^2 + (A + C + D)t + (B + D)$ Comparing coefficients, we obtain the following system of equations: \begin{align*} A + C &= 0 \ B + C + D &= 0 \ A + C + D &= 1 \ B + D &= 0 \end{align*} From the first equation, $C = -A$. From the fourth equation, $D = -B$. Substituting into the second and third equations, we get: \begin{align*} B - A - B &= 0 \implies -A = 0 \implies A = 0 \ 0 - A - B &= 1 \implies D = 1\end{align*} Since $A = 0$, $C = 0$. Since $D = 1$, $B = -1$. Thus, $\frac{t}{(1 + t + t^2)(1 + t^2)} = \frac{-1}{1 + t + t^2} + \frac{1}{1 + t^2}$

Step 3: Integration Now we integrate the decomposed expression: $\int \frac{t}{(1 + t + t^2)(1 + t^2)} dt = \int \left( \frac{1}{1 + t^2} - \frac{1}{1 + t + t^2} \right) dt$ $= \int \frac{1}{1 + t^2} dt - \int \frac{1}{t^2 + t + 1} dt$ The first integral is $\tan^{-1}(t)$. For the second integral, we complete the square in the denominator: $t^2 + t + 1 = t^2 + t + \frac{1}{4} + \frac{3}{4} = \left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$ So, the second integral becomes: $\int \frac{1}{\left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} dt = \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C_1 = \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t + 1}{\sqrt{3}}\right) + C_1$ Therefore, $\int \frac{t}{(1 + t + t^2)(1 + t^2)} dt = \tan^{-1}(t) - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2t + 1}{\sqrt{3}}\right) + C$ Substituting back $t = \tan x$, we have: $\int \frac{\tan x}{1 + \tan x + \tan^2 x} dx = x - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2\tan x + 1}{\sqrt{3}}\right) + C$

Step 4: Comparison with given form We are given that: $\int \frac{\tan x}{1 + \tan x + \tan^2 x} dx = x - \frac{K}{\sqrt{A}} \tan^{-1}\left(\frac{K \tan x + 1}{\sqrt{A}}\right) + C$ Comparing this with our result: $x - \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2\tan x + 1}{\sqrt{3}}\right) + C$ We can see that $K = 2$ and $A = 3$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs during partial fraction decomposition and when completing the square. A single sign error can invalidate the entire solution.
• Forgetting Constants: Remember to add the constant of integration, C, after each indefinite integral.
• Completing the Square: Practice completing the square quickly and accurately to handle quadratic expressions in the denominator.

Summary

We used trigonometric substitution to convert the integral into a rational function, then applied partial fraction decomposition to simplify it. After integrating each term, we completed the square to match the given form and finally compared coefficients to find the values of K and A. Therefore, the ordered pair (K, A) is (2, 3).

Final Answer The final answer is \boxed{(2, 3)}, which corresponds to option (C).