The integral shown in the image is:

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

Solution:

Step 1: Substitution

Let: $u = \sqrt{1 + x} \quad \Rightarrow \quad u^2 = 1 + x \quad \text{and} \quad x = u^2 - 1$

Differentiating $u^2 = 1 + x$: $2u \, du = dx$

Substituting into the integral:

$\int \frac{\sqrt{1 + x}}{1 - x} \, dx = \int \frac{u}{1 - (u^2 - 1)} \cdot 2u \, du$

Simplify the denominator $1 - (u^2 - 1)$: $1 - u^2 + 1 = 2 - u^2$

Thus, the integral becomes: $2 \int \frac{u^2}{2 - u^2} \, du$

Step 2: Simplify and Partial Fraction

Rewrite the fraction: $\frac{u^2}{2 - u^2} = -1 + \frac{2}{2 - u^2}$

Now the integral is split as: $2 \int \left( -1 + \frac{2}{2 - u^2} \right) \, du = -2 \int 1 \, du + 2 \int \frac{2}{2 - u^2} \, du$

Step 3: Solve Individual Terms

1. For $-2 \int 1 \, du$: $-2 \int 1 \, du = -2u$

2. For $2 \int \frac{1}{2 - u^2} \, du$, use the formula: $\frac{1}{a^2 - u^2} = \frac{1}{2a} \ln \left| \frac{a + u}{a - u} \right|$ Here $a^2 = 2$, so $a = \sqrt{2}$: $\int \frac{1}{2 - u^2} \, du = \frac{1}{2\sqrt{2}} \ln \left| \frac{\sqrt{2} + u}{\sqrt{2} - u} \right|$

Thus: $2 \int \frac{1}{2 - u^2} \, du = \frac{\sqrt{2}}{2} \ln \left| \frac{\sqrt{2} + u}{\sqrt{2} - u} \right|$

Step 4: Combine Results

Combine the terms: $\int \frac{\sqrt{1 + x}}{1 - x} \, dx = -2u + \sqrt{2} \ln \left| \frac{\sqrt{2} + u}{\sqrt{2} - u} \right| + C$

Substitute back $u = \sqrt{1 + x}$: $\int \frac{\sqrt{1 + x}}{1 - x} \, dx = -2\sqrt{1 + x} + \sqrt{2} \ln \left| \frac{\sqrt{2} + \sqrt{1 + x}}{\sqrt{2} - \sqrt{1 + x}} \right| + C$

Final Answer:

$\boxed{\int \frac{\sqrt{1 + x}}{1 - x} \, dx = -2\sqrt{1 + x} + \sqrt{2} \ln \left| \frac{\sqrt{2} + \sqrt{1 + x}}{\sqrt{2} - \sqrt{1 + x}} \right| + C}$

Would you like more details or a simpler explanation of any part of this?

Related Questions:

1. How do substitutions simplify complex integrals?
2. What is the formula for integrating functions with $a^2 - u^2$?
3. Why is logarithmic integration used in this problem?
4. How can partial fractions help solve similar integrals?
5. What are common applications of these integral techniques in physics or engineering?

Tip: Always check if substitutions reduce the integral into standard forms for faster solving!