`int (x^2-2)/(x^3 sqrt(x^2-1)) dx` is equal to

Options

A. `(x^(2))/(sqrt(x^(2)-1))+C`
B. `-(x^(2))/(sqrt(x^(2)-1))+C`
C. `(sqrt(x^(2)-1))/(x^(2))+C`
D. `-(sqrt(x^(2)-1))/(x^(2))+C`

Answer

To solve the integral

\[ I = \int \frac{x^2 - 2}{x^3 \sqrt{x^2 - 1}} \, dx, \]

we can start by rewriting the integrand. Notice that we can separate the integral into two parts:

\[ I = \int \frac{x^2}{x^3 \sqrt{x^2 - 1}} \, dx - \int \frac{2}{x^3 \sqrt{x^2 - 1}} \, dx. \]

This simplifies to:

\[ I = \int \frac{1}{x \sqrt{x^2 - 1}} \, dx - 2 \int \frac{1}{x^3 \sqrt{x^2 - 1}} \, dx. \]

### Step 1: Substitution
We will use the substitution \( u = \sqrt{x^2 - 1} \). Then, differentiating both sides gives:

\[ du = \frac{x}{\sqrt{x^2 - 1}} \, dx \quad \Rightarrow \quad dx = \frac{\sqrt{x^2 - 1}}{x} \, du. \]

From our substitution, we also have \( x^2 = u^2 + 1 \), which implies \( x = \sqrt{u^2 + 1} \).

### Step 2: Change of Variables
Now, we can express the integrals in terms of \( u \):

1. For the first integral:

\[ \int \frac{1}{x \sqrt{x^2 - 1}} \, dx = \int \frac{1}{\sqrt{u^2 + 1} \cdot u} \cdot \frac{u}{\sqrt{u^2}} \, du = \int \frac{1}{\sqrt{u^2 + 1}} \, du. \]

2. For the second integral:

\[ -2 \int \frac{1}{x^3 \sqrt{x^2 - 1}} \, dx = -2 \int \frac{1}{(u^2 + 1)^{3/2} \cdot u} \cdot \frac{u}{\sqrt{u^2}} \, du = -2 \int \frac{1}{(u^2 + 1)^{3/2}} \, du. \]

### Step 3: Evaluate the Integrals
Now we need to evaluate the two integrals:

1. The first integral:

\[ \int \frac{1}{\sqrt{u^2 + 1}} \, du = \ln |u + \sqrt{u^2 + 1}| + C_1. \]

2. The second integral:

Using the formula for the integral of \( \frac{1}{(u^2 + a^2)^{n}} \):

\[ -2 \int \frac{1}{(u^2 + 1)^{3/2}} \, du = -2 \cdot \frac{u}{(u^2 + 1)^{1/2}} + C_2. \]

### Step 4: Combine Results
Combining these results, we have:

\[ I = \ln |u + \sqrt{u^2 + 1}| - \frac{2u}{\sqrt{u^2 + 1}} + C. \]

### Step 5: Substitute Back
Now substituting back \( u = \sqrt{x^2 - 1} \):

\[ I = \ln |\sqrt{x^2 - 1} + x| - \frac{2\sqrt{x^2 - 1}}{x} + C. \]

### Final Answer
Thus, the integral evaluates to:

\[ \int \frac{x^2 - 2}{x^3 \sqrt{x^2 - 1}} \, dx = \ln |\sqrt{x^2 - 1} + x| - \frac{2\sqrt{x^2 - 1}}{x} + C. \]