Question:

Step 1: Perform a suitable substitution to simplify the integral. Let $t = x^2$. Differentiating both sides with respect to $x$, we get: $$dt = 2x \, dx$$ Step 2: Rewrite the integral in terms of $t$. Substitute $x^2 = t$ and $2x \, dx = dt$ into 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 integrand. The expression inside the square root is a quadratic in $t$. We complete the square to bring it to a standard form: $$ t^2 + 2t + 4 = (t^2 + 2t + 1) + 3 = (t+1)^2 + (\sqrt{3})^2 $$ 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 integral formula $\int \frac{du}{\sqrt{u^2 + a^2}} = \ln \left| u + \sqrt{u^2 + a^2} \right| + C$. Here, $u = t+1$ and $a = \sqrt{3}$. Applying the formula, we get: $$ \ln \left| (t+1) + \sqrt{(t+1)^2 + (\sqrt{3})^2} \right| + C $$ Step 5: Substitute back to express the result in terms of $x$. Replace $t$ with $x^2$ in the expression. Also, recall that $(t+1)^2 + (\sqrt{3})^2 = t^2 + 2t + 4$, which when $t=x^2$ becomes $x^4 + 2x^2 + 4$. $$ \ln \left| (x^2 + 1) + \sqrt{(x^2+1)^2 + (\sqrt{3})^2} \right| + C $$ $$ = \ln \left| (x^2 + 1) + \sqrt{x^4 + 2x^2 + 1 + 3} \right| + C $$ $$ = \ln \left| (x^2 + 1) + \sqrt{x^4 + 2x^2 + 4} \right| + C $$ Step 6: State the final answer. The evaluated integral is: $$ \int \frac{2x}{\sqrt{x^4 + 2x^2 + 4}} dx = \ln \left| x^2 + 1 + \sqrt{x^4 + 2x^2 + 4} \right| + C $$ This matches option A.