Evaluate: `inta/(b+c e^x)dx`

Answer

To evaluate the integral

\[ I = \int \frac{a}{b + c e^x} \, dx, \]

where \(a\), \(b\), and \(c\) are arbitrary constants, we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:

\[ I = \int \frac{a}{b + c e^x} \, dx. \]

### Step 2: Factor out \(e^x\)
We can factor \(e^x\) from the denominator:

\[ I = \int \frac{a e^{-x}}{b e^{-x} + c} \, dx. \]

### Step 3: Substitute
Let

\[ t = b e^{-x} + c. \]

Now, we differentiate \(t\) with respect to \(x\):

\[ \frac{dt}{dx} = -b e^{-x}. \]

Thus, we can express \(dx\) in terms of \(dt\):

\[ dx = \frac{dt}{-b e^{-x}}. \]

### Step 4: Substitute \(dx\) in the Integral
Now substituting \(dx\) in the integral:

\[ I = \int \frac{a e^{-x}}{t} \left(-\frac{dt}{b e^{-x}}\right). \]

This simplifies to:

\[ I = -\frac{a}{b} \int \frac{1}{t} \, dt. \]

### Step 5: Integrate
The integral of \(\frac{1}{t}\) is:

\[ \int \frac{1}{t} \, dt = \ln |t| + C. \]

Thus, we have:

\[ I = -\frac{a}{b} \ln |t| + C. \]

### Step 6: Substitute Back for \(t\)
Now we substitute back for \(t\):

\[ t = b e^{-x} + c. \]

So, we get:

\[ I = -\frac{a}{b} \ln |b e^{-x} + c| + C. \]

### Final Result
The final result of the integral is:

\[ I = -\frac{a}{b} \ln |b e^{-x} + c| + C. \]