Question

Evaluate: \[ \int \sqrt{x^2 + 2x + 5} \, dx \]

Hint:

To evaluate integrals of the form \( \int \sqrt{x^2 + a^2} \, dx \), use the standard formula and complete the square if necessary.

Answer 1

We begin by completing the square inside the square root: \[ x^2 + 2x + 5 = (x + 1)^2 + 4 \] So, the integral becomes: \[ \int \sqrt{(x + 1)^2 + 4} \, dx \] Now, use the substitution \( u = x + 1 \), so \( du = dx \). The integral becomes: \[ \int \sqrt{u^2 + 4} \, du \] This is a standard integral of the form \( \int \sqrt{u^2 + a^2} \, du \), which is: \[ \int \sqrt{u^2 + a^2} \, du = \frac{u}{2} \sqrt{u^2 + a^2} + \frac{a^2}{2} \ln\left(u + \sqrt{u^2 + a^2}\right) + C \] Substitute \( a = 2 \) and \( u = x + 1 \) back into the formula: \[ \int \sqrt{(x + 1)^2 + 4} \, dx = \frac{x + 1}{2} \sqrt{(x + 1)^2 + 4} + 2 \ln \left( x + 1 + \sqrt{(x + 1)^2 + 4} \right) + C \] Final Answer: \[ \boxed{\frac{x + 1}{2} \sqrt{(x + 1)^2 + 4} + 2 \ln \left( x + 1 + \sqrt{(x + 1)^2 + 4} \right) + C} \]