To solve the integral

$\int \frac{1}{\sqrt{x} + \sqrt{1 - x}} \, dx$

Let's go through the steps:

Step 1: Substitution

We introduce a substitution to simplify the expression: Let $u = \sqrt{x} + \sqrt{1 - x}$, which implies that $\frac{1}{u} = \frac{1}{\sqrt{x} + \sqrt{1 - x}}$.

Now, differentiate both sides with respect to $x$:

$\frac{du}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{1 - x}}$

From this, solving for $dx$ requires expressing this in terms of the new variable $u$, so we should now move on and explore ways to simplify this integral using trigonometric or more direct substitutions.

Step 2: Another Substitution (Trigonometric)

Let us substitute $x = \sin^2(\theta)$. Then:

$\sqrt{x} = \sin(\theta), \quad \sqrt{1 - x} = \cos(\theta)$

Thus, the integral becomes:

$\int \frac{1}{\sin(\theta) + \cos(\theta)} \cdot 2\sin(\theta)\cos(\theta) \, d\theta$

This can be simplified using trigonometric identities and integrating accordingly.

Let me compute this transformation and the final result for you.The integral in terms of the variable $\theta$ becomes:

$\frac{\sqrt{2}}{2} \left( \log\left(\tan\left(\frac{\theta}{2}\right) - 1 + \sqrt{2}\right) - \log\left(\tan\left(\frac{\theta}{2}\right) - \sqrt{2} - 1\right) \right)$

This is the result of transforming the integral using the substitution $x = \sin^2(\theta)$.

If you have any questions about this, or if you'd like to see how this result can be further interpreted or used in practice, feel free to ask!

Follow-up Questions:

1. How can this substitution method be applied to other integrals with square roots?
2. What other trigonometric substitutions are commonly useful in integrals?
3. How does the choice of substitution affect the difficulty of the problem?
4. Can the result of this integral be expressed in a simpler form?
5. What techniques can be used for integrals involving square roots and polynomials?

Tip: When faced with complicated integrals, trigonometric substitutions often simplify the problem by leveraging known identities.