Question

Evaluate : \( \int_0^2 \frac{1}{\sqrt{x^2 + 2x + 3}} \, dx \)

Hint:

Always complete the square when you see a quadratic under a square root in the denominator.
Check your standard integral formulas; confusing \( \sqrt{x^2 + a^2} \) with \( \sqrt{a^2 - x^2} \) is a very common error.

Answer 1

Concept:
• Completing the square: \( ax^2 + bx + c = a(x + \frac{b}{2a})^2 + (c - \frac{b^2}{4a}) \).
• Standard integral: \( \int \frac{dx}{\sqrt{x^2 + a^2}} = \log |x + \sqrt{x^2 + a^2}| + C \).

Step 1: Complete the square for the quadratic expression in the denominator
The quadratic is \( x^2 + 2x + 3 \). \[ x^2 + 2x + 3 = (x^2 + 2x + 1) + 2 = (x + 1)^2 + (\sqrt{2})^2 \]

Step 2: Rewrite the integral and apply the standard formula
Let \( I = \int_0^2 \frac{dx}{\sqrt{(x + 1)^2 + (\sqrt{2})^2}} \). Using the formula \( \int \frac{dx}{\sqrt{X^2 + A^2}} = \log |X + \sqrt{X^2 + A^2}| \), where \( X = x + 1 \) and \( A = \sqrt{2} \): \[ I = \left[ \log |(x + 1) + \sqrt{(x + 1)^2 + (\sqrt{2})^2}| \right]_0^2 \] Simplifying the term inside the square root back to the original quadratic: \[ I = \left[ \log |x + 1 + \sqrt{x^2 + 2x + 3}| \right]_0^2 \]

Step 3: Evaluate the limits
Upper limit (\( x = 2 \)): \[ \log |2 + 1 + \sqrt{2^2 + 2(2) + 3}| = \log |3 + \sqrt{4 + 4 + 3}| = \log(3 + \sqrt{11}) \] Lower limit (\( x = 0 \)): \[ \log |0 + 1 + \sqrt{0^2 + 2(0) + 3}| = \log |1 + \sqrt{3}| = \log(1 + \sqrt{3}) \] Subtracting the results: \[ I = \log(3 + \sqrt{11}) - \log(1 + \sqrt{3}) \] Using the property \( \log a - \log b = \log(a/b) \): \[ I = \log \left( \frac{3 + \sqrt{11}}{1 + \sqrt{3}} \right) \]