Find $\int \sqrt{x^2 + 2x + 5} \, dx$ |
$\frac{x + 1}{2}\sqrt{x^2 + 2x + 5} + 2 \ln |x + 1 + \sqrt{x^2 + 2x + 5}| + C$ $(x + 1)\sqrt{x^2 + 2x + 5} + 4 \ln |x + 1 + \sqrt{x^2 + 2x + 5}| + C$ $\frac{x + 1}{2}\sqrt{x^2 + 2x + 5} - 2 \ln |x + 1 + \sqrt{x^2 + 2x + 5}| + C$ $\frac{x}{2}\sqrt{x^2 + 2x + 5} + 2 \ln |x + \sqrt{x^2 + 2x + 5}| + C$ |
$\frac{x + 1}{2}\sqrt{x^2 + 2x + 5} + 2 \ln |x + 1 + \sqrt{x^2 + 2x + 5}| + C$ |
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: $\int \sqrt{x^2 + 2x + 5} \, dx = \int \sqrt{y^2 + 2^2} \, dy \text{}$ $= \frac{1}{2} y \sqrt{y^2 + 4} + \frac{4}{2} \log |y + \sqrt{y^2 + 4}| + C$ $= \frac{1}{2} (x+1) \sqrt{x^2 + 2x + 5} + 2 \log |x+1 + \sqrt{x^2 + 2x + 5}| + C \text{}$ |