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

Options

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

Answer

To solve the integral \(\int \frac{dx}{x(x+1)}\), we can use the method of partial fractions. Here are the steps to solve it:

### Step 1: Decompose into Partial Fractions
We start by expressing the integrand \(\frac{1}{x(x+1)}\) as a sum of partial fractions:
\[ \frac{1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \]
where \(A\) and \(B\) are constants that we need to determine.

### Step 2: Find Constants A and B
Multiplying both sides by \(x(x+1)\) gives:
\[ 1 = A(x+1) + Bx \]
Expanding the right side:
\[ 1 = Ax + A + Bx \]
Combining like terms:
\[ 1 = (A + B)x + A \]
Now, we can set up the equations by comparing coefficients:
1. \(A + B = 0\) (coefficient of \(x\))
2. \(A = 1\) (constant term)

From the second equation, we find \(A = 1\). Substituting \(A\) into the first equation:
\[ 1 + B = 0 \implies B = -1 \]

### Step 3: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \frac{dx}{x(x+1)} = \int \left(\frac{1}{x} - \frac{1}{x+1}\right) dx \]

### Step 4: Integrate Each Term
Now we integrate each term separately:
\[ \int \left(\frac{1}{x} - \frac{1}{x+1}\right) dx = \int \frac{1}{x} dx - \int \frac{1}{x+1} dx \]
The integrals are:
\[ \int \frac{1}{x} dx = \log |x| + C_1 \]
\[ \int \frac{1}{x+1} dx = \log |x+1| + C_2 \]
Thus, combining these results:
\[ \int \frac{dx}{x(x+1)} = \log |x| - \log |x+1| + C \]

### Step 5: Use Logarithmic Properties
Using the property of logarithms \(\log a - \log b = \log \frac{a}{b}\):
\[ \log |x| - \log |x+1| = \log \left|\frac{x}{x+1}\right| \]
Thus, the final result is:
\[ \int \frac{dx}{x(x+1)} = \log \left|\frac{x}{x+1}\right| + C \]

### Final Answer
\[ \int \frac{dx}{x(x+1)} = \log \left|\frac{x}{x+1}\right| + C \]

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