If h(a)=h(b), the value of the integral
`int_a^b [f(g(h(x))]^(-1)f'(g(h(x)))g'(h(x))h'(x)dx`is equal to

Options

A. 0
B. f(a)-f(b)
C. f(g(a))-f(g(b))
D. none of these

Answer

To solve the integral

\[ I = \int_a^b \left[ f(g(h(x))) \right]^{-1} f'(g(h(x))) g'(h(x)) h'(x) \, dx \]

given that \( h(a) = h(b) \), we can follow these steps:

### Step 1: Substitution
Let \( t = f(g(h(x))) \). Then we need to find \( dt \).

**Hint:** Use the chain rule for differentiation.

### Step 2: Differentiate \( t \)
Using the chain rule, we have:

\[ dt = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \, dx \]

Rearranging gives us:

\[ dx = \frac{dt}{f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)} \]

**Hint:** Remember that \( dx \) can be expressed in terms of \( dt \).

### Step 3: Substitute in the Integral
Now substitute \( t \) and \( dx \) into the integral:

\[ I = \int_{h(a)}^{h(b)} \frac{1}{t} dt \]

**Hint:** Change the limits of integration according to the substitution.

### Step 4: Evaluate the Integral
The integral of \( \frac{1}{t} \) is:

\[ I = \left[ \log |t| \right]_{h(a)}^{h(b)} \]

**Hint:** Remember to apply the limits correctly.

### Step 5: Substitute Back
Now substitute back \( t = f(g(h(x))) \):

\[ I = \log(f(g(h(b)))) - \log(f(g(h(a)))) \]

**Hint:** Use properties of logarithms to simplify.

### Step 6: Apply the Given Condition
Since \( h(a) = h(b) \), we have:

\[ I = \log(f(g(h(a)))) - \log(f(g(h(a)))) \]

This simplifies to:

\[ I = \log(f(g(h(a)))) - \log(f(g(h(a)))) = 0 \]

**Hint:** The logarithm of a number minus itself is always zero.

### Final Result
Thus, the value of the integral is:

\[ \boxed{0} \]