Certainly! Let's solve the integral $\int \frac{x+1}{x^2+2x+19} \, dx$ using substitution.

Step 1: Recognize the structure

The integrand has the form $\frac{x + 1}{x^2 + 2x + 19}$. Notice that the derivative of the denominator $x^2 + 2x + 19$ is $2x + 2$, which is close to the numerator $x + 1$. This suggests that a substitution involving the denominator might simplify the integral.

Step 2: Complete the square for the denominator

First, complete the square for the quadratic expression in the denominator: $x^2 + 2x + 19 = (x + 1)^2 + 18$ So the integral becomes: $\int \frac{x+1}{(x+1)^2 + 18} \, dx$

Step 3: Use substitution

Let $u = x + 1$. Then, $du = dx$. The integral now becomes: $\int \frac{u}{u^2 + 18} \, du$

Step 4: Split the integral

The integral can be split into two parts: $\int \frac{u}{u^2 + 18} \, du = \frac{1}{2} \int \frac{2u \, du}{u^2 + 18}$ Notice that the derivative of $u^2 + 18$ is $2u$. Therefore, the integral simplifies to: $\frac{1}{2} \int \frac{d(u^2 + 18)}{u^2 + 18}$

This is a standard logarithmic form: $\frac{1}{2} \ln|u^2 + 18| + C$

Step 5: Substitute back $u = x + 1$

Substituting back $u = x + 1$, we get: $\frac{1}{2} \ln\left((x + 1)^2 + 18\right) + C$

This simplifies to: $\boxed{\frac{1}{2} \ln\left(x^2 + 2x + 19\right) + C}$

So, the final answer is: $\boxed{\frac{1}{2} \ln\left(x^2 + 2x + 19\right) + C}$

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. What would the solution be if the numerator was $x + 3$ instead of $x + 1$?
2. How would you approach this integral if the quadratic in the denominator were more complex?
3. Can you solve a similar integral using a different method, like partial fractions?
4. What happens if we apply substitution directly on $x^2 + 2x + 19$ without completing the square?
5. How can you verify the result by differentiating the solution?

Tip: Whenever you encounter a quadratic in the denominator, completing the square is a helpful step to simplify the integral, especially when combined with substitution.