If `y(x-y)^2=x`, then `int1/(x-3y)dx` is equal to (A) `1/3log{(x-y)^2+1}` (B) `1/4log{(x-y)^2-1}` (C) `1/2log{(x-y)^2-1}` (D) `1/6 log{(x^2-y^2-1}`

Options

A. 1
B. 3
C. 5
D. 7

Answer

To solve the problem where \( y(x - y)^2 = x \) and we need to find \( \int \frac{1}{x - 3y} \, dx \), we can follow these steps:

### Step 1: Differentiate the given equation
We start with the equation:
\[ y(x - y)^2 = x \]
Differentiating both sides with respect to \( x \) using the product rule:
\[ \frac{d}{dx}[y(x - y)^2] = \frac{d}{dx}[x] \]
Applying the product rule:
\[ y \cdot 2(x - y)(1 - \frac{dy}{dx}) + (x - y)^2 \frac{dy}{dx} = 1 \]

### Step 2: Rearranging the differentiated equation
Rearranging the equation gives:
\[ y \cdot 2(x - y)(1 - y') + (x - y)^2 y' = 1 \]
Where \( y' = \frac{dy}{dx} \).

### Step 3: Collect terms involving \( y' \)
Factor out \( y' \):
\[ y'[(x - y)^2 + 2y(x - y)] = 1 - 2y(x - y) \]
Thus,
\[ y' = \frac{1 - 2y(x - y)}{(x - y)^2 + 2y(x - y)} \]

### Step 4: Find \( x - 3y \)
From the differentiated equation, we can express \( x - 3y \) in terms of \( y \):
\[ x - 3y = (x - y) - 2y \]

### Step 5: Rewrite the integral
We need to find:
\[ \int \frac{1}{x - 3y} \, dx \]
Using the expression for \( x - 3y \):
\[ \int \frac{1}{(x - y) - 2y} \, dx \]

### Step 6: Substitute and integrate
Let \( t = (x - y)^2 - 1 \), then:
\[ dt = 2(x - y) \, dx \]
Thus,
\[ dx = \frac{dt}{2(x - y)} \]
Substituting into the integral:
\[ \int \frac{1}{x - 3y} \, dx = \int \frac{1}{(x - y) - 2y} \cdot \frac{dt}{2(x - y)} \]

### Step 7: Solve the integral
This leads to:
\[ \frac{1}{2} \int \frac{1}{t} \, dt = \frac{1}{2} \log |t| + C \]
Substituting back for \( t \):
\[ = \frac{1}{2} \log |(x - y)^2 - 1| + C \]

### Final Answer
Thus, the final result is:
\[ \int \frac{1}{x - 3y} \, dx = \frac{1}{2} \log |(x - y)^2 - 1| + C \]