The value of `int((x-4))/(x^2sqrt(x-2))` dx is equal to (where , C is the constant of integration )

Options

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

Answer

To solve the integral \( \int \frac{x-4}{x^2 \sqrt{x-2}} \, dx \), we will use a substitution method. Here’s the step-by-step solution:

### Step 1: Substitution
Let \( x - 2 = t^2 \). Then, we have:
\[ x = t^2 + 2 \]
Differentiating both sides gives:
\[ dx = 2t \, dt \]

### Step 2: Rewrite the Integral
Now, we can express \( x - 4 \) and \( \sqrt{x - 2} \) in terms of \( t \):
\[ x - 4 = (t^2 + 2) - 4 = t^2 - 2 \]
\[ \sqrt{x - 2} = \sqrt{t^2} = t \]
Thus, the integral becomes:
\[ \int \frac{t^2 - 2}{(t^2 + 2)^2} \cdot 2t \, dt \]
This simplifies to:
\[ 2 \int \frac{t(t^2 - 2)}{(t^2 + 2)^2} \, dt \]

### Step 3: Simplify the Integral
Now we can break this down:
\[ 2 \int \frac{t^3 - 2t}{(t^2 + 2)^2} \, dt \]
This can be split into two separate integrals:
\[ 2 \left( \int \frac{t^3}{(t^2 + 2)^2} \, dt - 2 \int \frac{t}{(t^2 + 2)^2} \, dt \right) \]

### Step 4: Solve Each Integral
1. **For \( \int \frac{t^3}{(t^2 + 2)^2} \, dt \)**:
Use the substitution \( u = t^2 + 2 \), then \( du = 2t \, dt \) or \( dt = \frac{du}{2t} \):
\[ \int \frac{t^3}{(t^2 + 2)^2} \, dt = \int \frac{t^2}{u^2} \cdot \frac{du}{2t} = \frac{1}{2} \int \frac{t^2}{u^2} \, du \]
Since \( t^2 = u - 2 \):
\[ = \frac{1}{2} \int \frac{u - 2}{u^2} \, du = \frac{1}{2} \left( \int \frac{1}{u} \, du - 2 \int \frac{1}{u^2} \, du \right) \]

2. **For \( \int \frac{t}{(t^2 + 2)^2} \, dt \)**:
Again use the substitution \( u = t^2 + 2 \):
\[ = \frac{1}{2} \int \frac{1}{u^2} \, du \]

### Step 5: Combine Results
After integrating, we will combine the results and substitute back \( t = \sqrt{x - 2} \).

### Final Result
After performing the integrations and substituting back, we find:
\[ \int \frac{x-4}{x^2 \sqrt{x-2}} \, dx = -\frac{2\sqrt{x-2}}{x} + C \]