To solve the integral [tex]\(\int \frac{dx}{\sqrt{x^2 + x - 2}}\)[/tex], let's go through the steps to find the result:
Step 1: Start with the given integral:
[tex]\[
\int \frac{dx}{\sqrt{x^2 + x - 2}}
\][/tex]
Step 2: We need to simplify the expression inside the square root. Let's complete the square for the quadratic expression [tex]\(x^2 + x - 2\)[/tex].
First, rewrite [tex]\(x^2 + x - 2\)[/tex] in a form suitable for completing the square:
[tex]\[
x^2 + x - 2 = x^2 + x + \frac{1}{4} - \frac{1}{4} - 2 = \left(x + \frac{1}{2}\right)^2 - \frac{9}{4}
\][/tex]
Now the integral becomes:
[tex]\[
\int \frac{dx}{\sqrt{\left(x + \frac{1}{2}\right)^2 - \left(\frac{3}{2}\right)^2}}
\][/tex]
Step 3: Let [tex]\(u = x + \frac{1}{2}\)[/tex]. Then [tex]\(du = dx\)[/tex] and the integral changes to:
[tex]\[
\int \frac{du}{\sqrt{u^2 - \left(\frac{3}{2}\right)^2}}
\][/tex]
Step 4: Recognize the structure of the integrand. The integral [tex]\(\int \frac{du}{\sqrt{u^2 - a^2}}\)[/tex] is a standard integral and its result is [tex]\(\ln|u + \sqrt{u^2 - a^2}| + C\)[/tex], where [tex]\(a\)[/tex] is a constant.
In our case, [tex]\(a = \frac{3}{2}\)[/tex], so the integrand becomes:
[tex]\[
\int \frac{du}{\sqrt{u^2 - \left(\frac{3}{2}\right)^2}}
\][/tex]
Using the standard form:
[tex]\[
\int \frac{du}{\sqrt{u^2 - \left(\frac{3}{2}\right)^2}} = \ln\left|u + \sqrt{u^2 - \left(\frac{3}{2}\right)^2}\right| + C
\][/tex]
Step 5: Substitute back the value of [tex]\(u\)[/tex]:
[tex]\[
u = x + \frac{1}{2}
\][/tex]
So the integral becomes:
[tex]\[
\ln\left|x + \frac{1}{2} + \sqrt{\left(x + \frac{1}{2}\right)^2 - \left(\frac{3}{2}\right)^2}\right| + C
\][/tex]
Step 6: Finally, simplify the expression inside the logarithm:
[tex]\[
\ln\left|x + \frac{1}{2} + \sqrt{x^2 + x - 2}\right| + C
\][/tex]
Since we ignore the absolute value for a cleaner result, we can express the final answer as:
[tex]\[
\boxed{\ln\left(2x + 2\sqrt{x^2 + x - 2} + 1\right) + C}
\][/tex]
Therefore, the result of the integral is:
[tex]\[
\ln\left(2x + 2\sqrt{x^2 + x - 2} + 1\right) + C
\][/tex]
