`int(x-1)/(x+1)^(2)dx=`

Options

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

Answer

To solve the integral \( \int \frac{x-1}{(x+1)^2} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting the integrand \( \frac{x-1}{(x+1)^2} \) by adding and subtracting 1 in the numerator:

\[ \frac{x-1}{(x+1)^2} = \frac{(x+1) - 2}{(x+1)^2} = \frac{x+1}{(x+1)^2} - \frac{2}{(x+1)^2} \]

### Step 2: Split the integral
Now we can split the integral into two separate integrals:

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

### Step 3: Simplify the first integral
The first integral simplifies as follows:

\[ \int \frac{x+1}{(x+1)^2} \, dx = \int \frac{1}{x+1} \, dx \]

### Step 4: Integrate the first term
The integral of \( \frac{1}{x+1} \) is:

\[ \int \frac{1}{x+1} \, dx = \ln |x+1| \]

### Step 5: Integrate the second term
Now we integrate the second term:

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

The integral of \( \frac{1}{(x+1)^2} \) is:

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

Thus, we have:

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

### Step 6: Combine the results
Now we can combine the results from the two integrals:

\[ \int \frac{x-1}{(x+1)^2} \, dx = \ln |x+1| + \frac{2}{x+1} + C \]

where \( C \) is the constant of integration.

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

\[ \int \frac{x-1}{(x+1)^2} \, dx = \ln |x+1| + \frac{2}{x+1} + C \]