`int(1)/(x^(2)-x^(3))dx=`

Options

A. `log((1-x)/(x))-(1)/(x)+`
B. `log((1-x)/(x))+(1)/(x)+c`
C. `log((x)/(1+x))+c`
D. `-log((1-x)/(x))-(1)/(x)+c`

Answer

To solve the integral \( \int \frac{1}{x^2 - x^3} \, dx \), we can follow these steps:

### Step 1: Simplify the Denominator
First, we can factor the denominator:

\[ x^2 - x^3 = x^2(1 - x) \]

Thus, the integral becomes:

\[ \int \frac{1}{x^2(1 - x)} \, dx \]

### Step 2: Rewrite the Integral
We can rewrite the integral as:

\[ \int \frac{1}{x^2} \cdot \frac{1}{1 - x} \, dx \]

### Step 3: Use Substitution
Let’s use the substitution \( t = \frac{1}{x} - 1 \). Then, we have:

\[ \frac{1}{x} = t + 1 \implies x = \frac{1}{t + 1} \]

Now we differentiate \( x \):

\[ dx = -\frac{1}{(t + 1)^2} dt \]

### Step 4: Substitute in the Integral
Now we substitute \( x \) and \( dx \) into the integral:

\[ \int \frac{1}{\left(\frac{1}{t + 1}\right)^2} \cdot \frac{1}{1 - \frac{1}{t + 1}} \left(-\frac{1}{(t + 1)^2}\right) dt \]

This simplifies to:

\[ \int \frac{(t + 1)^2}{1 - \frac{1}{t + 1}} \left(-\frac{1}{(t + 1)^2}\right) dt \]

### Step 5: Simplify Further
The term \( 1 - \frac{1}{t + 1} \) simplifies to \( \frac{t}{t + 1} \). So now we have:

\[ -\int \frac{1}{\frac{t}{t + 1}} dt = -\int \frac{t + 1}{t} dt \]

### Step 6: Break Down the Integral
This can be split into two integrals:

\[ -\int \left(1 + \frac{1}{t}\right) dt = -\left(t + \log |t|\right) + C \]

### Step 7: Substitute Back
Now we substitute back for \( t \):

\[ t = \frac{1}{x} - 1 \implies -\left(\frac{1}{x} - 1 + \log \left|\frac{1}{x} - 1\right|\right) + C \]

### Step 8: Final Simplification
This leads us to:

\[ \log \left| \frac{1}{x} - 1 \right| + 1 + C \]

Thus, the final answer is:

\[ -\log \left| \frac{1 - x}{x} \right| + C \]