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.