`int(cosx-1)/(sinx+1).e^(x)dx` is equal to

Options

A. `(e^(x)cosx)/(1+sinx)+C`
B. `C-(e^(x)sinx)/(1+sinx)`
C. `C-(e^(x))/(1+sinx)`
D. `C-(e^(x)cosx)/(1+sinx)`

Explanation

To solve the integral \( I = \int \frac{\cos x - 1}{\sin x + 1} e^x \, dx \), we can break it down into manageable parts. Here’s the step-by-step solution:

### Step 1: Rewrite the Integral
We start with the given integral:
\[ I = \int \frac{\cos x - 1}{\sin x + 1} e^x \, dx \]
We can separate the integral into two parts:
\[ I = \int \frac{\cos x}{\sin x + 1} e^x \, dx - \int \frac{1}{\sin x + 1} e^x \, dx \]

### Step 2: Apply Integration by Parts
For the first integral \( \int \frac{\cos x}{\sin x + 1} e^x \, dx \), we will use integration by parts. Let:
- \( u = \frac{\cos x}{\sin x + 1} \)
- \( dv = e^x \, dx \)

Then we find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \left( \frac{-\sin x (\sin x + 1) - \cos x \cos x}{(\sin x + 1)^2} \right) dx = \frac{-\sin x (\sin x + 1) - \cos^2 x}{(\sin x + 1)^2} \, dx \]
- Integrate \( dv \):
\[ v = e^x \]

### Step 3: Apply the Integration by Parts Formula
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I_1 = \frac{\cos x}{\sin x + 1} e^x - \int e^x \left( \frac{-\sin x (\sin x + 1) - \cos^2 x}{(\sin x + 1)^2} \right) \, dx \]

### Step 4: Simplify the Integral
Now we simplify the integral:
\[ I_1 = \frac{\cos x}{\sin x + 1} e^x + \int \frac{e^x (\sin x (\sin x + 1) + \cos^2 x)}{(\sin x + 1)^2} \, dx \]

### Step 5: Combine the Integrals
Now we can combine both parts of the integral:
\[ I = I_1 - I_2 \]
Where \( I_2 = \int \frac{e^x}{\sin x + 1} \, dx \).

### Step 6: Final Result
After evaluating both integrals and simplifying, we find:
\[ I = \frac{e^x \cos x}{\sin x + 1} + C \]
where \( C \) is the constant of integration.

### Final Answer
Thus, the integral is:
\[ \int \frac{\cos x - 1}{\sin x + 1} e^x \, dx = \frac{e^x \cos x}{\sin x + 1} + C \]