Let's solve the given integral:

$I = \int_{0}^{1} \frac{x}{(x+1)(x^2 + 2x + 1)} \,dx$

Step 1: Factorizing the Denominator

The denominator can be rewritten as:

$(x+1)(x^2 + 2x + 1) = (x+1)(x+1)^2 = (x+1)^3$

So, the integral simplifies to:

$I = \int_{0}^{1} \frac{x}{(x+1)^3} \,dx$

Step 2: Substituting $u = x+1$

Let:

$u = x+1 \Rightarrow du = dx$

When $x = 0$, then $u = 1$, and when $x = 1$, then $u = 2$. Thus, the integral transforms into:

$I = \int_{1}^{2} \frac{u-1}{u^3} \,du$

Step 3: Splitting the Fraction

Rewriting:

$\frac{u-1}{u^3} = \frac{u}{u^3} - \frac{1}{u^3} = \frac{1}{u^2} - \frac{1}{u^3}$

Thus, the integral becomes:

$I = \int_{1}^{2} \left( u^{-2} - u^{-3} \right) du$

Step 4: Integrating

Using the power rule for integration:

$\int u^n \,du = \frac{u^{n+1}}{n+1}, \quad \text{for } n \neq -1$

We compute:

$\int u^{-2} \,du = \frac{u^{-1}}{-1} = -\frac{1}{u}$

$\int u^{-3} \,du = \frac{u^{-2}}{-2} = -\frac{1}{2u^2}$

Thus:

$I = \left[ -\frac{1}{u} + \frac{1}{2u^2} \right]_{1}^{2}$

Step 5: Evaluating at Limits

Substituting $u = 2$:

$-\frac{1}{2} + \frac{1}{2(2^2)} = -\frac{1}{2} + \frac{1}{8} = -\frac{4}{8} + \frac{1}{8} = -\frac{3}{8}$

Substituting $u = 1$:

$-\frac{1}{1} + \frac{1}{2(1^2)} = -1 + \frac{1}{2} = -\frac{2}{2} + \frac{1}{2} = -\frac{1}{2}$

Thus:

$I = \left(-\frac{3}{8} \right) - \left(-\frac{1}{2} \right)$

$I = -\frac{3}{8} + \frac{4}{8} = \frac{1}{8}$

Final Answer:

$\frac{1}{8}$

Would you like a more detailed explanation or a verification?

Here are five related questions you might find interesting:

1. How do you perform substitution in definite integrals?
2. What are common techniques for evaluating rational function integrals?
3. How does partial fraction decomposition help in integration?
4. What is the fundamental theorem of calculus and how does it apply here?
5. Can we evaluate this integral using an alternative approach?

Tip: When integrating rational functions, always check for possible factorizations before applying advanced techniques!