To integrate the function [tex]\(\left(x^2 + x + \frac{1}{2}\right)^{-1/2}\)[/tex], let's proceed step-by-step.
1. Identify the integral:
We need to evaluate the integral:
[tex]\[
\int \left(x^2 + x + \frac{1}{2}\right)^{-1/2} \, dx
\][/tex]
2. Recognize the form:
Notice that the integrand [tex]\(\left(x^2 + x + \frac{1}{2}\right)^{-1/2}\)[/tex] has a quadratic expression inside the parentheses.
3. Complete the square:
Before proceeding, we should rewrite the quadratic expression [tex]\(x^2 + x + \frac{1}{2}\)[/tex] in a more convenient form by completing the square.
[tex]\[
x^2 + x + \frac{1}{2} = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{1}{2} = \left(x + \frac{1}{2}\right)^2 + \frac{1}{4}
\][/tex]
Thus, the integrand becomes:
[tex]\[
\int \left(\left(x + \frac{1}{2}\right)^2 + \frac{1}{4}\right)^{-1/2} \, dx
\][/tex]
4. Substitution:
Let [tex]\(u = x + \frac{1}{2}\)[/tex], then [tex]\(du = dx\)[/tex]. The integral changes to:
[tex]\[
\int \left(u^2 + \frac{1}{4}\right)^{-1/2} \, du
\][/tex]
5. Recognize a standard integral:
The integral [tex]\(\int \left(u^2 + a^2\right)^{-1/2} \, du\)[/tex] where [tex]\(a\)[/tex] is a constant, is a standard integral. By looking up standard forms, we know that:
[tex]\[
\int \left(u^2 + a^2\right)^{-1/2} \, du = \text{arsinh}\left(\frac{u}{a}\right) + C
\][/tex]
For [tex]\(a = \frac{1}{2}\)[/tex], it modifies to:
[tex]\[
\int \left(u^2 + \left(\frac{1}{2}\right)^2\right)^{-1/2} \, du = \text{arsinh}(2u) + C
\][/tex]
6. Back-substitute for [tex]\(u\)[/tex]:
Recall that [tex]\(u = x + \frac{1}{2}\)[/tex]. Substituting back [tex]\(u\)[/tex]:
[tex]\[
\int \left(x^2 + x + \frac{1}{2}\right)^{-1/2} \, dx = \text{arsinh}\left(2\left(x + \frac{1}{2}\right)\right) + C
\][/tex]
7. Simplify the final result:
[tex]\[
2\left(x + \frac{1}{2}\right) = 2x + 1
\][/tex]
Thus, our integral evaluates to:
[tex]\[
\text{arsinh}(2x + 1) + C
\][/tex]
So, the result of the integral [tex]\(\int \left(x^2 + x + \frac{1}{2}\right)^{-1/2} \, dx\)[/tex] is:
[tex]\[
\boxed{\text{arsinh}(2x + 1) + C}
\][/tex]
