The equation you've provided is the integral of:

$\int \frac{dx}{\sqrt{1 - x - x^2}}$

To solve this integral, we can use a substitution technique to simplify the expression. Let's go through the steps:

1. Recognize the structure: The denominator resembles a quadratic expression under the square root, so we can attempt a substitution to reduce it into a more solvable form.

2. Substitution: Let's complete the square on the quadratic expression $1 - x - x^2$.

$1 - x - x^2 = 1 - (x^2 + x)$

We complete the square for $x^2 + x$:

$x^2 + x = \left(x + \frac{1}{2}\right)^2 - \frac{1}{4}$

Thus,

$1 - x - x^2 = 1 - \left[\left(x + \frac{1}{2}\right)^2 - \frac{1}{4}\right] = \frac{5}{4} - \left(x + \frac{1}{2}\right)^2$

So, the integral becomes:

$\int \frac{dx}{\sqrt{\frac{5}{4} - \left(x + \frac{1}{2}\right)^2}}$

3. Further substitution: Let $u = x + \frac{1}{2}$. Then $du = dx$, and the integral transforms into:

$\int \frac{du}{\sqrt{\frac{5}{4} - u^2}}$

4. Solve using a trigonometric substitution: Let $u = \frac{\sqrt{5}}{2} \sin \theta$, then $du = \frac{\sqrt{5}}{2} \cos \theta d\theta$, and the integral becomes:

$\int \frac{\frac{\sqrt{5}}{2} \cos \theta d\theta}{\sqrt{\frac{5}{4} - \frac{5}{4} \sin^2 \theta}} = \int d\theta = \theta + C$

5. Back-substitute: Since $\theta = \arcsin\left(\frac{2u}{\sqrt{5}}\right)$, we back-substitute $u = x + \frac{1}{2}$:

$\theta = \arcsin\left(\frac{2(x + \frac{1}{2})}{\sqrt{5}}\right)$

Thus, the final solution to the integral is:

$\arcsin\left(\frac{2(x + \frac{1}{2})}{\sqrt{5}}\right) + C$

Conclusion:

The solution to the integral is:

$\arcsin\left(\frac{2(x + \frac{1}{2})}{\sqrt{5}}\right) + C$

Would you like more details or have any questions on the steps?

Here are 5 related questions to expand on this topic:

1. How do you complete the square for different quadratic expressions?
2. Why is trigonometric substitution useful in integrals involving square roots?
3. Can this integral be solved using any other substitution methods?
4. How do you handle constants in integrals after substitution?
5. What are the common techniques to solve integrals involving quadratics under square roots?

Tip: Completing the square is a powerful technique that helps transform complex quadratic expressions into simpler forms, making them easier to integrate.