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

Options

A. `(1)/(2)log|x+1|+(1)/(2(x+1))-(1)/(4)log|x^2+1|+c`
B. `(1)/(2)log|x+1|-(1)/(2(x+1))+(1)/(4)log|x^2+1|+c`
C. `(1)/(2)log|x+1|-(1)/(2(x+1))-(1)/(4)log|x^2+1|+c`
D. `(1)/(2)log|x+1|+(1)/(2(x+1))+(1)/(4)log|x^2+1|+c`

Answer

To evaluate the integral

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

we will use the method of partial fractions.

### Step 1: Set up the partial fraction decomposition

We can express the integrand as:

\[ \frac{1}{(x+1)^2 (x^2+1)} = \frac{Ax + B}{(x+1)^2} + \frac{Cx + D}{x^2 + 1}. \]

### Step 2: Clear the denominators

Multiplying both sides by \((x+1)^2 (x^2+1)\) gives:

\[ 1 = (Ax + B)(x^2 + 1) + (Cx + D)(x + 1)^2. \]

### Step 3: Expand the right-hand side

Expanding both terms:

1. \((Ax + B)(x^2 + 1) = Ax^3 + Bx^2 + Ax + B\)
2. \((Cx + D)(x + 1)^2 = (Cx + D)(x^2 + 2x + 1) = Cx^3 + (2C + D)x^2 + (C + 2D)x + D\)

Combining these, we have:

\[ 1 = (A + C)x^3 + (B + 2C + D)x^2 + (A + C + 2D)x + (B + D). \]

### Step 4: Set up the system of equations

Equating the coefficients from both sides gives us the following system of equations:

1. \(A + C = 0\) (coefficient of \(x^3\))
2. \(B + 2C + D = 0\) (coefficient of \(x^2\))
3. \(A + C + 2D = 0\) (coefficient of \(x\))
4. \(B + D = 1\) (constant term)

### Step 5: Solve the system of equations

From equation 1, we can express \(C\) in terms of \(A\):

\[ C = -A. \]

Substituting \(C = -A\) into the other equations:

- From equation 2:
\[ B + 2(-A) + D = 0 \implies B - 2A + D = 0 \implies D = 2A - B. \]

- From equation 3:
\[ A - A + 2D = 0 \implies 2D = 0 \implies D = 0. \]

Substituting \(D = 0\) into the equation for \(B\):

\[ B + 0 = 1 \implies B = 1. \]

Now substituting \(B = 1\) back into \(D = 2A - B\):

\[ D = 2A - 1 = 0 \implies 2A = 1 \implies A = \frac{1}{2}. \]

Now substituting \(A = \frac{1}{2}\) into \(C = -A\):

\[ C = -\frac{1}{2}. \]

### Step 6: Summary of coefficients

We have found:

- \(A = \frac{1}{2}\)
- \(B = 1\)
- \(C = -\frac{1}{2}\)
- \(D = 0\)

Thus, the partial fraction decomposition is:

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

### Step 7: Integrate each term

Now we can integrate each term separately:

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

#### Integrating the first term

Using substitution \(u = x + 1\), \(du = dx\):

\[ \int \frac{\frac{1}{2}(u - 1) + 1}{u^2} \, du = \int \frac{\frac{1}{2}u + \frac{1}{2}}{u^2} \, du = \frac{1}{2} \int \frac{1}{u} \, du + \frac{1}{2} \int \frac{1}{u^2} \, du. \]

This results in:

\[ \frac{1}{2} \ln |u| - \frac{1}{2u} + C_1 = \frac{1}{2} \ln |x + 1| - \frac{1}{2(x + 1)}. \]

#### Integrating the second term

Using substitution \(v = x^2 + 1\), \(dv = 2x \, dx\):

\[ -\frac{1}{2} \int \frac{1}{v} \, dv = -\frac{1}{2} \ln |v| + C_2 = -\frac{1}{2} \ln |x^2 + 1|. \]

### Step 8: Combine results

Combining the results, we have:

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

### Final Answer

Thus, the final answer is:

\[ \frac{1}{2} \ln |x + 1| - \frac{1}{2(x + 1)} - \frac{1}{2} \ln |x^2 + 1| + C. \]