Find ∫ √(x^2 + 2x + 5) dx

The correct answer is Option (1) → $\frac{x + 1}{2}\sqrt{x^2 + 2x + 5} + 2 \ln |x + 1 + \sqrt{x^2 + 2x + 5}| + C$

$\int \sqrt{x^2 + 2x + 5} \, dx = \int \sqrt{(x+1)^2 + 4} \, dx$.

Put $x + 1 = y$, so that $dx = dy$. Then:

$\int \sqrt{x^2 + 2x + 5} \, dx = \int \sqrt{y^2 + 2^2} \, dy \text{}$

$= \frac{1}{2} y \sqrt{y^2 + 4} + \frac{4}{2} \log |y + \sqrt{y^2 + 4}| + C$

$= \frac{1}{2} (x+1) \sqrt{x^2 + 2x + 5} + 2 \log |x+1 + \sqrt{x^2 + 2x + 5}| + C \text{}$