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.

Learn how to solve the integral ∫(x+1)/(x^2+2x+19) dx step-by-step with a detailed solution using partial fractions and substitution methods. This problem involves integration techniques and algebraic manipulation. Suitable for undergraduate students studying calculus and integration.

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