To solve the given integral

$\int_1^2 \frac{2x^2 \, dx}{(x^3 + 1)^2} = \frac{a}{27},$

where $a = ?$, we will proceed as follows:

Step 1: Substitution

Let $u = x^3 + 1$. Then, $du = 3x^2 dx$. Notice that $x^2 dx = \frac{1}{3} du$.

When $x = 1$, $u = 1^3 + 1 = 2$.

When $x = 2$, $u = 2^3 + 1 = 9$.

Thus, the integral becomes: $\int_1^2 \frac{2x^2 \, dx}{(x^3 + 1)^2} = \int_2^9 \frac{2}{u^2} \cdot \frac{1}{3} \, du.$

Step 2: Simplify the new integral

Simplify the constants: $\int_2^9 \frac{2}{u^2} \cdot \frac{1}{3} \, du = \frac{2}{3} \int_2^9 u^{-2} \, du.$

Step 3: Solve the integral

The integral of $u^{-2}$ is: $\int u^{-2} \, du = -u^{-1} = -\frac{1}{u}.$

Thus: $\frac{2}{3} \int_2^9 u^{-2} \, du = \frac{2}{3} \left[ -\frac{1}{u} \right]_2^9.$

Step 4: Evaluate the limits

Evaluate $\left[ -\frac{1}{u} \right]_2^9$: $\left[ -\frac{1}{u} \right]_2^9 = -\frac{1}{9} - \left(-\frac{1}{2}\right) = -\frac{1}{9} + \frac{1}{2}.$

Simplify: $-\frac{1}{9} + \frac{1}{2} = \frac{-2 + 9}{18} = \frac{7}{18}.$

Thus: $\frac{2}{3} \cdot \frac{7}{18} = \frac{14}{54} = \frac{7}{27}.$

Step 5: Identify $a$

The result is: $\int_1^2 \frac{2x^2 \, dx}{(x^3 + 1)^2} = \frac{7}{27}.$

Comparing with $\frac{a}{27}$, we find: $a = 7.$

Final Answer:

$a = 7.$

Would you like me to explain any step further or clarify concepts? 😊

Related Questions:

1. How does substitution simplify integrals involving composite functions?
2. What is the importance of adjusting limits when performing substitution in definite integrals?
3. How do constants factor out in integrals, and how do they affect the result?
4. Why does integrating $u^{-2}$ yield $-\frac{1}{u}$?
5. Can this integral be solved without substitution, and how would it differ?

Tip:

Always double-check the limits of integration when using substitution, as failing to adjust them correctly is a common mistake.