Question:

<p><strong>Step 1:</strong> Rewrite the expression under the square root as a sum of squares.</p><p><span class="math">$x^2 + 4x + 1 = (x+2)^2 - 3$</span></p><p><strong>Step 2:</strong> Use substitution <span class="math">$u = x + 2$</span>, so <span class="math">$du = dx$</span></p><p><span class="math">$\int \sqrt{u^2 - 3}\,du$</span></p><p><strong>Step 3:</strong> Apply the standard formula <span class="math">$\int \sqrt{t^2 - a^2}\,dt = \frac{t}{2}\sqrt{t^2-a^2} - \frac{a^2}{2}\log|t + \sqrt{t^2-a^2}| + C$</span></p><p><span class="math">$= \frac{u}{2}\sqrt{u^2-3} - \frac{3}{2}\log|u + \sqrt{u^2-3}| + C$</span></p><p><strong>Step 4:</strong> Substitute back <span class="math">$u = x+2$</span></p><p><span class="math">$= \frac{(x+2)}{2}\sqrt{x^2+4x+1} - \frac{3}{2}\log|(x+2) + \sqrt{x^2+4x+1}| + C$</span></p><p>∴ Answer is A.</p>