The value of `int(logx)/(x+1)^(2)dx` is

Options

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

Answer

To solve the integral \( \int \frac{\log x}{(x+1)^2} \, dx \), we will use integration by parts. Let's denote:

- \( u = \log x \)
- \( dv = \frac{1}{(x+1)^2} \, dx \)

### Step 1: Differentiate \( u \) and Integrate \( dv \)

First, we differentiate \( u \) and integrate \( dv \):

- \( du = \frac{1}{x} \, dx \)
- To find \( v \), we integrate \( dv \):

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

Using the integral formula \( \int \frac{1}{a^2} \, dx = -\frac{1}{a} + C \), we get:

\[ v = -\frac{1}{x+1} \]

### Step 2: Apply Integration by Parts Formula

Using the integration by parts formula:

\[ \int u \, dv = uv - \int v \, du \]

we substitute our values:

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

This simplifies to:

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

### Step 3: Simplify the Remaining Integral

Now, we need to evaluate \( \int \frac{1}{x(x+1)} \, dx \). We can use partial fraction decomposition:

\[ \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \]

Multiplying through by \( x(x+1) \):

\[ 1 = A(x+1) + Bx \]

Setting \( x = 0 \):

\[ 1 = A(0+1) \Rightarrow A = 1 \]

Setting \( x = -1 \):

\[ 1 = B(-1) \Rightarrow B = -1 \]

Thus, we have:

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

### Step 4: Integrate the Partial Fractions

Now we can integrate:

\[ \int \frac{1}{x(x+1)} \, dx = \int \left(\frac{1}{x} - \frac{1}{x+1}\right) \, dx = \log |x| - \log |x+1| + C \]

### Step 5: Combine Everything

Now we substitute back into our expression:

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

This simplifies to:

\[ -\frac{\log x}{x+1} + \log \left|\frac{x}{x+1}\right| + C \]

### Final Answer

Thus, the final result is:

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