Question:

Note that <br> $\int \sqrt{x^2 + 2x + 5} dx = \int \sqrt{(x+1)^2 + 4} dx$ <br> Put $x + 1 = y$, so that $dx = dy$. Then <br> $\int \sqrt{x^2 + 2x + 5} dx = \int \sqrt{y^2 + 2^2} dy = \frac{1}{2}y\sqrt{y^2 + 4} + \frac{4}{2}\log|y + \sqrt{y^2 + 4}| + C$ <br> $= \frac{1}{2}(x+1)\sqrt{x^2 + 2x + 5} + 2\log|x+1 + \sqrt{x^2 + 2x + 5}| + C$