To solve the integral [tex]\(\int \frac{1}{x^2 + 10x + 30} \, dx\)[/tex], we can follow these steps:
1. Complete the square for the quadratic expression in the denominator:
[tex]\[
x^2 + 10x + 30
\][/tex]
To complete the square, we need to express [tex]\(x^2 + 10x + 30\)[/tex] in the form [tex]\((x + a)^2 + b\)[/tex].
Start by focusing on the quadratic and linear terms:
[tex]\[
x^2 + 10x
\][/tex]
Add and subtract [tex]\(\left(\frac{10}{2}\right)^2 = 25\)[/tex]:
[tex]\[
x^2 + 10x + 25 - 25 + 30
\][/tex]
This can be rewritten as:
[tex]\[
(x + 5)^2 + 5
\][/tex]
2. Substitute the completed square into the integral:
[tex]\[
\int \frac{1}{(x + 5)^2 + 5} \, dx
\][/tex]
3. Use a trigonometric substitution to simplify the integral. Let [tex]\(u = x + 5\)[/tex], hence [tex]\(du = dx\)[/tex]. The integral now becomes:
[tex]\[
\int \frac{1}{u^2 + 5} \, du
\][/tex]
4. Factor the constant in the denominator to fit a standard integral form:
[tex]\[
\int \frac{1}{u^2 + (\sqrt{5})^2} \, du
\][/tex]
Here we recognize the integrand as the form [tex]\(\frac{1}{a^2 + u^2}\)[/tex], whose antiderivative is [tex]\(\frac{1}{a} \arctan \left(\frac{u}{a}\right)\)[/tex].
5. Integrate using the standard formula:
[tex]\[
\int \frac{1}{u^2 + (\sqrt{5})^2} \, du = \frac{1}{\sqrt{5}} \arctan \left(\frac{u}{\sqrt{5}}\right) + C
\][/tex]
6. Substitute back [tex]\(u = x + 5\)[/tex] to return to the original variable [tex]\(x\)[/tex]:
[tex]\[
\frac{1}{\sqrt{5}} \arctan \left(\frac{x + 5}{\sqrt{5}}\right) + C
\][/tex]
Therefore, the integral [tex]\(\int \frac{1}{x^2 + 10x + 30} \, dx\)[/tex] evaluates to:
[tex]\[
\frac{1}{\sqrt{5}} \arctan \left(\frac{x + 5}{\sqrt{5}}\right) + C
\][/tex]
Given the provided steps and results, this is the antiderivative of the given integral.
