`int(x-1)/((x+1)^(3))e^(x)dx=`

Options

A. `(-e^(x))/((x+1)^(2))+c`
B. `(e^(x))/(x+1)+c`
C. `(-e^(x))/((x+1)^(2))+c`
D. `(e^(x))/(x-1)+c`

Answer

To solve the integral

\[ \int \frac{x-1}{(x+1)^3} e^x \, dx, \]

we can utilize the property of integration involving the product of an exponential function and the derivative of a function.

### Step 1: Rewrite the integrand
We start by rewriting the integrand to facilitate the use of the integration property. We can express \(x - 1\) as follows:

\[ x - 1 = (x + 1) - 2. \]

So, we can rewrite the integral as:

\[ \int \frac{(x + 1) - 2}{(x + 1)^3} e^x \, dx = \int \left( \frac{x + 1}{(x + 1)^3} - \frac{2}{(x + 1)^3} \right) e^x \, dx. \]

### Step 2: Separate the integrals
Now we can separate the integral into two parts:

\[ \int \frac{1}{(x + 1)^2} e^x \, dx - 2 \int \frac{1}{(x + 1)^3} e^x \, dx. \]

### Step 3: Apply integration by parts
For the first integral, we can use the integration property mentioned in the transcript. Let:

\[ f(x) = \frac{1}{x + 1} \implies f'(x) = -\frac{1}{(x + 1)^2}. \]

Thus, we can express the first integral as:

\[ \int e^x f'(x) \, dx = e^x f(x) + C = e^x \cdot \frac{1}{x + 1} + C. \]

### Step 4: Solve the second integral
For the second integral, we can similarly use the property. Let:

\[ g(x) = \frac{1}{(x + 1)^2} \implies g'(x) = -\frac{2}{(x + 1)^3}. \]

Thus, we can express the second integral as:

\[ \int e^x g'(x) \, dx = e^x g(x) + C = e^x \cdot \frac{1}{(x + 1)^2} + C. \]

### Step 5: Combine the results
Now we can combine the results of both integrals:

\[ \int \frac{x - 1}{(x + 1)^3} e^x \, dx = e^x \cdot \frac{1}{x + 1} - 2 \left( e^x \cdot \frac{1}{(x + 1)^2} \right) + C. \]

### Final Answer
Thus, the final answer is:

\[ \int \frac{x - 1}{(x + 1)^3} e^x \, dx = e^x \left( \frac{1}{x + 1} - \frac{2}{(x + 1)^2} \right) + C. \]