Question

Evaluate the integral: \[ I = \int \frac{2}{1 + x + x^2} \,dx. \]

• A. \( \frac{4}{\sqrt{3}} \tan^{-1} \left(\frac{2x - 1}{\sqrt{3}}\right) + c \)
• B. \( \frac{4}{\sqrt{3}} \tan^{-1} \left(\frac{2x + 1}{\sqrt{3}}\right) + c \)
• C. \( \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2x - 1}{\sqrt{3}}\right) + c \)
• D. \( \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2x + 1}{\sqrt{3}}\right) + c \)

Hint:

For integrals of the form \( \int \frac{dx}{x^2 + bx + c} \), complete the square in the denominator and use the standard inverse tangent integral formula.

Answer 1

The Correct Option is B

Step 1: Completing the square in the denominator We start with the denominator: \[ 1 + x + x^2. \] Rearrange the terms: \[ x^2 + x + 1. \] To complete the square, write: \[ x^2 + x + \frac{1}{4} + \frac{3}{4}. \] \[ = \left(x + \frac{1}{2} \right)^2 + \frac{3}{4}. \]
Step 2: Using substitution Let: \[ t = x + \frac{1}{2}. \] Then: \[ dt = dx. \] The denominator transforms into: \[ t^2 + \frac{3}{4}. \] Thus, our integral becomes: \[ I = \int \frac{2}{t^2 + \frac{3}{4}} \, dt. \] Using the standard formula: \[ \int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right), \] where \( a^2 = \frac{3}{4} \Rightarrow a = \frac{\sqrt{3}}{2} \), we obtain: \[ I = \frac{4}{\sqrt{3}} \tan^{-1} \left( \frac{2t}{\sqrt{3}} \right) + C. \] Substituting \( t = x + \frac{1}{2} \): \[ I = \frac{4}{\sqrt{3}} \tan^{-1} \left( \frac{2(x + \frac{1}{2})}{\sqrt{3}} \right) + C. \] \[ = \frac{4}{\sqrt{3}} \tan^{-1} \left( \frac{2x + 1}{\sqrt{3}} \right) + C. \] % Final Answer Thus, the correct answer is option (2).