To solve the integral $\int \frac{x+1}{x^2+2x+19} \, dx$, we can proceed by breaking it down using partial fractions or by using a substitution method.

Step 1: Simplify the integrand

We can rewrite the denominator $x^2 + 2x + 19$ as: $x^2 + 2x + 19 = (x+1)^2 + 18$ So, the integral becomes: $\int \frac{x+1}{(x+1)^2 + 18} \, dx$

Step 2: Substitution

Let's use the substitution $u = x + 1$, so $du = dx$. The integral becomes: $\int \frac{du}{u^2 + 18}$

Step 3: Solve the integral

This is a standard integral form: $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C$ where $a = \sqrt{18} = 3\sqrt{2}$. Therefore, the integral becomes: $\frac{1}{3\sqrt{2}} \arctan\left(\frac{u}{3\sqrt{2}}\right) + C$

Step 4: Substitute back $u = x + 1$

Substituting back $u = x + 1$: $\int \frac{x+1}{x^2+2x+19} \, dx = \frac{1}{3\sqrt{2}} \arctan\left(\frac{x+1}{3\sqrt{2}}\right) + C$

So, the final answer is: $\boxed{\frac{1}{3\sqrt{2}} \arctan\left(\frac{x+1}{3\sqrt{2}}\right) + C}$

Would you like more details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the integral change if the numerator was $x - 1$ instead of $x + 1$?
2. What if the quadratic in the denominator had a different coefficient for the $x^2$ term?
3. Can you solve a similar integral if the quadratic is factorizable?
4. How does the presence of different coefficients in the quadratic affect the arctangent form?
5. What if the integral were a definite integral, say from 0 to 1?

Tip: When dealing with quadratic expressions in the denominator, completing the square is a powerful technique to simplify the integration process.