If the integral `I=int(x sqrtx-3x+3sqrtx-1)/(x-2sqrtx+1)dx=f(x)+C` (where, `x gt0` and C is the constant of integration) and `f(1)=(-1)/(3)`, then the value of `f(9)` is equal to

Options

A. 3
B. 6
C. 9
D. 12

Answer

To solve the integral

\[ I = \int \frac{x \sqrt{x} - 3x + 3\sqrt{x} - 1}{x - 2\sqrt{x} + 1} \, dx = f(x) + C \]

with the condition \( f(1) = -\frac{1}{3} \), we need to find \( f(9) \).

### Step 1: Substitution
Let's perform the substitution \( t = \sqrt{x} \). Then, we have:

\[ x = t^2 \quad \text{and} \quad dx = 2t \, dt \]

### Step 2: Rewrite the Integral
Substituting \( x \) and \( dx \) into the integral gives:

\[ I = \int \frac{t^2 t - 3t^2 + 3t - 1}{t^2 - 2t + 1} \cdot 2t \, dt \]

This simplifies to:

\[ I = 2 \int \frac{t^3 - 3t^2 + 3t - 1}{(t-1)^2} \, dt \]

### Step 3: Simplify the Numerator
The numerator can be factored or simplified. Notice that:

\[ t^3 - 3t^2 + 3t - 1 = (t-1)^3 \]

Thus, we can rewrite the integral as:

\[ I = 2 \int \frac{(t-1)^3}{(t-1)^2} \, dt = 2 \int (t-1) \, dt \]

### Step 4: Integrate
Now we can integrate:

\[ I = 2 \left( \frac{(t-1)^2}{2} \right) + C = (t-1)^2 + C \]

### Step 5: Back Substitute
Substituting back \( t = \sqrt{x} \):

\[ I = (\sqrt{x} - 1)^2 + C \]

### Step 6: Find \( f(x) \)
Thus, we have:

\[ f(x) = (\sqrt{x} - 1)^2 + C \]

### Step 7: Use the Condition \( f(1) = -\frac{1}{3} \)
Now we use the condition \( f(1) = -\frac{1}{3} \):

\[ f(1) = (\sqrt{1} - 1)^2 + C = 0 + C = C \]

So,

\[ C = -\frac{1}{3} \]

### Step 8: Final Expression for \( f(x) \)
Now substituting \( C \) back into \( f(x) \):

\[ f(x) = (\sqrt{x} - 1)^2 - \frac{1}{3} \]

### Step 9: Find \( f(9) \)
Now we can find \( f(9) \):

\[ f(9) = (\sqrt{9} - 1)^2 - \frac{1}{3} = (3 - 1)^2 - \frac{1}{3} = 2^2 - \frac{1}{3} = 4 - \frac{1}{3} = \frac{12}{3} - \frac{1}{3} = \frac{11}{3} \]

### Conclusion
Thus, the value of \( f(9) \) is:

\[ \boxed{\frac{11}{3}} \]