Question

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

• A. \( \frac{1}{2} \sqrt{x^2+x+1} + \frac{1}{2} \cosh^{-1} \left(\frac{x+2}{\sqrt{3}}\right) + C \)
• B. \( \frac{1}{2} \sqrt{x^2+x+1} + \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2x+1}{\sqrt{3}}\right) + C \)
• C. \( \sqrt{x^2+x+1} + \frac{2}{\sqrt{3}} \log |x^2 + x + 1| + C \)
• D. \( \sqrt{x^2+x+1} + \frac{1}{2} \sinh^{-1} \left(\frac{2x+1}{\sqrt{3}}\right) + C \)

Hint:

For integrals of the form \( \int \frac{x + c}{\sqrt{x^2 + ax + b}} dx \), try completing the square and using inverse hyperbolic functions.

Answer 1

The Correct Option is D

Step 1: Completing the square The denominator can be rewritten by completing the square: \[ x^2 + x + 1 = \left( x + \frac{1}{2} \right)^2 + \frac{3}{4}. \] Let \( u = x^2 + x + 1 \), then: \[ du = (2x+1) dx. \] Step 2: Splitting the integral Rewriting the given integral, \[ I = \int \frac{x+1}{\sqrt{x^2+x+1}} dx. \] Using substitution \( u = x^2 + x + 1 \), and separating terms, \[ I = \int \frac{(2x+1)}{2\sqrt{u}} dx + \int \frac{dx}{\sqrt{u}}. \] The first integral simplifies to \( \sqrt{u} \), and the second integral is evaluated using inverse hyperbolic functions: \[ \int \frac{dx}{\sqrt{x^2 + x + 1}} = \sinh^{-1} \left(\frac{2x+1}{\sqrt{3}}\right). \] Step 3: Final expression Thus, the final integral evaluates to: \[ I = \sqrt{x^2+x+1} + \frac{1}{2} \sinh^{-1} \left(\frac{2x+1}{\sqrt{3}}\right) + C. \]