Question:

Step 1: Introduce a suitable substitution to simplify the integral. We observe that the numerator contains $2x \, dx$ and the denominator contains terms involving $x^2$ and $x^4$. This suggests a substitution involving $x^2$. Let $t = x^2$. Differentiating both sides with respect to $x$, we get: $$ \frac{dt}{dx} = 2x $$ Rearranging, we find: $$ dt = 2x \, dx $$ Step 2: Substitute the new variable into the integral. Replace $x^2$ with $t$ and $2x \, dx$ with $dt$ in the given integral: $$ \int \frac{2x}{\sqrt{x^4 + 2x^2 + 4}} dx = \int \frac{dt}{\sqrt{t^2 + 2t + 4}} $$ Step 3: Complete the square in the denominator of the transformed integral. To integrate the expression $\int \frac{dt}{\sqrt{t^2 + 2t + 4}}$, we complete the square for the quadratic expression in the denominator, $t^2 + 2t + 4$. $$ t^2 + 2t + 4 = (t^2 + 2t + 1) + 3 = (t+1)^2 + (\sqrt{3})^2 $$ Now the integral becomes: $$ \int \frac{dt}{\sqrt{(t+1)^2 + (\sqrt{3})^2}} $$ Step 4: Apply the standard integration formula. We use the standard integration formula for integrals of the form $\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln \left| x + \sqrt{x^2 + a^2} \right| + C$. In our case, $x$ is replaced by $(t+1)$ and $a$ is replaced by $\sqrt{3}$. $$ \int \frac{dt}{\sqrt{(t+1)^2 + (\sqrt{3})^2}} = \ln \left| (t+1) + \sqrt{(t+1)^2 + (\sqrt{3})^2} \right| + C $$ Step 5: Substitute back the original variable. Substitute $t = x^2$ back into the result: $$ \ln \left| (x^2+1) + \sqrt{(x^2+1)^2 + (\sqrt{3})^2} \right| + C $$ Simplify the expression under the square root: $$ (x^2+1)^2 + (\sqrt{3})^2 = x^4 + 2x^2 + 1 + 3 = x^4 + 2x^2 + 4 $$ So, the integral becomes: $$ \ln \left| (x^2+1) + \sqrt{x^4 + 2x^2 + 4} \right| + C $$ Step 6: State the final answer. The evaluated integral is $\ln \left| (x^2+1) + \sqrt{x^4 + 2x^2 + 4} \right| + C$. This matches with option A. The final answer is $\boxed{\ln \left| (x^2+1) + \sqrt{x^4 + 2x^2 + 4} \right| + C}$.