`"If " f(x)=sqrt(x),g(x)=e^(x)-1,`and `int fog(x)dx=A fog(x)+B tan^(-1)(fog(x))+C,` then `A+B " is equal to ".`

Explanation

To solve the problem, we need to find the value of \( A + B \) given the functions \( f(x) = \sqrt{x} \) and \( g(x) = e^x - 1 \). We are also given the integral:

\[ \int f(g(x)) \, dx = A f(g(x)) + B \tan^{-1}(f(g(x))) + C \]

### Step-by-Step Solution:

1. **Find \( f(g(x)) \)**:
\[ g(x) = e^x - 1 \implies f(g(x)) = f(e^x - 1) = \sqrt{e^x - 1} \]

2. **Set up the integral**:
We need to evaluate:
\[ \int \sqrt{e^x - 1} \, dx \]

3. **Substitution**:
Let \( t = \sqrt{e^x - 1} \). Then, squaring both sides gives:
\[ t^2 = e^x - 1 \implies e^x = t^2 + 1 \]
Taking the natural logarithm:
\[ x = \ln(t^2 + 1) \]

4. **Differentiate to find \( dx \)**:
Differentiate \( e^x \):
\[ \frac{d}{dx}(e^x) = e^x \implies dx = \frac{2t}{e^x} dt \]
Substitute \( e^x = t^2 + 1 \):
\[ dx = \frac{2t}{t^2 + 1} dt \]

5. **Substitute into the integral**:
Now substituting \( \sqrt{e^x - 1} \) and \( dx \):
\[ \int \sqrt{e^x - 1} \, dx = \int t \cdot \frac{2t}{t^2 + 1} dt = 2 \int \frac{t^2}{t^2 + 1} dt \]

6. **Simplify the integral**:
Rewrite the integrand:
\[ \frac{t^2}{t^2 + 1} = 1 - \frac{1}{t^2 + 1} \]
Thus:
\[ 2 \int \left( 1 - \frac{1}{t^2 + 1} \right) dt = 2 \left( \int dt - \int \frac{1}{t^2 + 1} dt \right) \]

7. **Integrate**:
The integrals become:
\[ 2 \left( t - \tan^{-1}(t) \right) + C \]

8. **Back substitute \( t \)**:
Recall \( t = \sqrt{e^x - 1} \):
\[ 2 \left( \sqrt{e^x - 1} - \tan^{-1}(\sqrt{e^x - 1}) \right) + C \]

9. **Compare with the given form**:
We have:
\[ \int f(g(x)) \, dx = 2 \sqrt{e^x - 1} - 2 \tan^{-1}(\sqrt{e^x - 1}) + C \]
Comparing with:
\[ A f(g(x)) + B \tan^{-1}(f(g(x))) + C \]
We find:
- \( A = 2 \)
- \( B = -2 \)

10. **Calculate \( A + B \)**:
\[ A + B = 2 - 2 = 0 \]

### Final Answer:
\[ A + B = 0 \]