If `intf(x)dx=g(x)+c and f^(-1)(x)` is differentiable, then `intf^(-1)(x)dx` equal to

Options

A. `g^(-1)(x)+C`
B. `xf^(-1)+C`
C. `xf^(-1)(x)-g(f^(-1)(x))+C`
D. `f^(-1)(x)+C`

Explanation

To solve the problem, we need to find the integral of the inverse function \( f^{-1}(x) \) given that:

\[ \int f(x) \, dx = g(x) + c \]

where \( f^{-1}(x) \) is differentiable.

### Step-by-Step Solution:

1. **Set Up the Integral**:
We start with the integral we want to evaluate:
\[ \int f^{-1}(x) \, dx \]

2. **Use Integration by Parts**:
We can apply integration by parts, which states:
\[ \int u \, dv = uv - \int v \, du \]
Here, we let:
- \( u = f^{-1}(x) \)
- \( dv = dx \)

Thus, \( du = \frac{d}{dx} f^{-1}(x) \, dx \) and \( v = x \).

3. **Apply Integration by Parts**:
Now substituting into the integration by parts formula:
\[ \int f^{-1}(x) \, dx = f^{-1}(x) \cdot x - \int x \cdot \frac{d}{dx} f^{-1}(x) \, dx \]

4. **Change of Variables**:
To simplify the integral, we can use a substitution. Let:
\[ f^{-1}(x) = t \implies x = f(t) \]
Then, differentiating both sides gives:
\[ \frac{dx}{dt} = f'(t) \implies dx = f'(t) \, dt \]

5. **Substituting Back**:
Substitute \( x = f(t) \) into the integral:
\[ \int f^{-1}(x) \, dx = t \cdot f(t) - \int f(t) \cdot \frac{1}{f'(t)} \, dt \]

6. **Using the Given Information**:
From the problem, we know:
\[ \int f(t) \, dt = g(t) + c \]
Therefore, we can replace the integral:
\[ \int f(t) \cdot \frac{1}{f'(t)} \, dt = g(t) + c \]

7. **Final Expression**:
Putting everything together, we have:
\[ \int f^{-1}(x) \, dx = f^{-1}(x) \cdot x - (g(f^{-1}(x)) + c) \]
Thus, the final result is:
\[ \int f^{-1}(x) \, dx = x f^{-1}(x) - g(f^{-1}(x)) - c \]