To solve the integral [tex]\(\int \left( \frac{1}{x^2 + 10x + 30} \right) dx\)[/tex], follow these steps:
1. Complete the square: First, we write the quadratic expression [tex]\(x^2 + 10x + 30\)[/tex] in a form that reveals a perfect square plus a constant. Completing the square involves finding a constant such that we can reframe the quadratic expression. Here’s how:
[tex]\[
x^2 + 10x + 30 = (x^2 + 10x + 25) + 5 = (x + 5)^2 + 5
\][/tex]
2. Rewrite the integral: Substituting the completed square form into the integral,
[tex]\[
\int \left( \frac{1}{(x + 5)^2 + 5} \right) dx
\][/tex]
3. Use a substitution: To simplify the integral, we use the substitution [tex]\(u = x + 5\)[/tex]. Thus, [tex]\(du = dx\)[/tex]. Rewriting the integral in terms of [tex]\(u\)[/tex],
[tex]\[
\int \left( \frac{1}{u^2 + 5} \right) du
\][/tex]
4. Factor out constants: Note that [tex]\(5\)[/tex] is a constant, and we can factor it out to put the integral into a recognizable form:
[tex]\[
\int \left( \frac{1}{u^2 + \sqrt{5}^2} \right) du
\][/tex]
5. Recognize the standard form: The integral now matches the standard form of the arctangent function:
[tex]\[
\int \left( \frac{1}{a^2 + u^2} \right) du = \frac{1}{a} \arctan \left( \frac{u}{a} \right) + C
\][/tex]
Here, [tex]\(a = \sqrt{5}\)[/tex].
6. Apply the arctangent integral formula:
[tex]\[
\int \left( \frac{1}{u^2 + 5} \right) du = \frac{1}{\sqrt{5}} \arctan \left( \frac{u}{\sqrt{5}} \right) + C
\][/tex]
7. Substitute back [tex]\(u = x + 5\)[/tex] into the formula:
[tex]\[
\frac{1}{\sqrt{5}} \arctan \left( \frac{x + 5}{\sqrt{5}} \right) + C
\][/tex]
After performing these steps, we get to our final result:
[tex]\[
\boxed{\frac{\sqrt{5}}{5} \arctan\left(\frac{\sqrt{5}x}{5} + \sqrt{5}\right)}
\][/tex]
