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).

Answer 2

To solve the integral \( I = \int \frac{2}{1 + x + x^2} \,dx \), we need to simplify the denominator. We recognize that it can be rewritten by completing the square:

\( 1 + x + x^2 = x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \).

Substituting this back, we get:

\( I = \int \frac{2}{\left(x + \frac{1}{2}\right)^2 + \frac{3}{4}} \, dx \).

Recognizing this integral as an inverse tangent form, we use the substitution:

\( u = x + \frac{1}{2}, \quad du = dx \).

Then the integral becomes:

\( I = \int \frac{2}{u^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \, du \).

Using the standard formula \( \int \frac{1}{u^2 + a^2} \, du = \frac{1}{a} \tan^{-1} \left(\frac{u}{a}\right) + C \), where \( a = \frac{\sqrt{3}}{2} \), we find:

\( I = \frac{2}{\frac{\sqrt{3}}{2}} \tan^{-1} \left(\frac{u}{\frac{\sqrt{3}}{2}}\right) + C = \frac{4}{\sqrt{3}} \tan^{-1} \left(\frac{2u}{\sqrt{3}}\right) + C \).

Substituting back \( u = x + \frac{1}{2} \), the solution is:

\( \frac{4}{\sqrt{3}} \tan^{-1} \left(\frac{2(x + \frac{1}{2})}{\sqrt{3}}\right) + C = \frac{4}{\sqrt{3}} \tan^{-1} \left(\frac{2x + 1}{\sqrt{3}}\right) + C \).

Thus, the correct answer is:

\( \frac{4}{\sqrt{3}} \tan^{-1} \left(\frac{2x + 1}{\sqrt{3}}\right) + c \)