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The integral to solve is:

\[
\int \frac{5x + 7}{x^2 + 4x + 4} \, dx
\]

### Solution:
We can solve this integral using the method of **completing the square** and then applying a substitution.

1. **Complete the square** for the quadratic expression in the denominator:

   \[
   x^2 + 4x + 4 = (x + 2)^2
   \]

   So the integral becomes:

   \[
   \int \frac{5x + 7}{(x + 2)^2} \, dx
   \]

2. **Substitute** \( u = x + 2 \), hence \( du = dx \) and \( x = u - 2 \). 

   Substituting into the integral:

   \[
   \int \frac{5(u - 2) + 7}{u^2} \, du
   \]

   Simplify the numerator:

   \[
   \int \frac{5u - 10 + 7}{u^2} \, du = \int \frac{5u - 3}{u^2} \, du
   \]

   Break this into two integrals:

   \[
   \int \frac{5u}{u^2} \, du - \int \frac{3}{u^2} \, du
   \]

   Simplify each term:

   \[
   \int \frac{5}{u} \, du - \int 3u^{-2} \, du
   \]

3. **Integrate** each term:

   - The first term: \( \int \frac{5}{u} \, du = 5 \ln|u| \)
   - The second term: \( \int 3u^{-2} \, du = 3 \cdot \left(-u^{-1}\right) = -\frac{3}{u} \)

   So the integral becomes:

   \[
   5 \ln|u| + \frac{3}{u} + C
   \]

4. **Substitute back** \( u = x + 2 \):

   \[
   5 \ln|x + 2| + \frac{3}{x + 2} + C
   \]

Thus, the final answer is:

\[
\int \frac{5x + 7}{x^2 + 4x + 4} \, dx = 5 \ln|x + 2| + \frac{3}{x + 2} + C
\]

Would you like further details or have any questions?

Here are some related questions for further practice:
1. How would you solve \(\int \frac{3x + 5}{x^2 + 6x + 9} \, dx\)?
2. What is the integral of \(\int \frac{7x + 4}{x^2 + 2x + 1} \, dx\)?
3. How do you complete the square for the expression \(x^2 + 6x + 9\)?
4. What method would you use to solve \(\int \frac{2x + 3}{x^2 + 4} \, dx\)?
5. How does substitution help in solving complex integrals?

**Tip:** When dealing with rational functions, consider using substitution and completing the square to simplify the integrals.

Explore how to solve the integral of (5x + 7) / (x^2 + 4x + 4) step-by-step. Learn integration techniques including completing the square and substitution. Suitable for advanced high school students and anyone studying calculus.

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