`int(dx)/((16x^(2)-25))`

Answer

To solve the integral \(\int \frac{dx}{16x^2 - 25}\), we can follow these steps:

### Step 1: Factor the Denominator
First, we rewrite the denominator \(16x^2 - 25\) as a difference of squares:
\[ 16x^2 - 25 = (4x)^2 - (5)^2 = (4x - 5)(4x + 5) \]

### Step 2: Rewrite the Integral
Now we can express the integral as:
\[ \int \frac{dx}{(4x - 5)(4x + 5)} \]

### Step 3: Use Partial Fraction Decomposition
We can use partial fraction decomposition to express \(\frac{1}{(4x - 5)(4x + 5)}\) in a simpler form:
\[ \frac{1}{(4x - 5)(4x + 5)} = \frac{A}{4x - 5} + \frac{B}{4x + 5} \]
Multiplying through by the denominator \((4x - 5)(4x + 5)\) gives:
\[ 1 = A(4x + 5) + B(4x - 5) \]

### Step 4: Solve for A and B
Expanding the right side:
\[ 1 = (4A + 4B)x + (5A - 5B) \]
Setting the coefficients equal gives us the system of equations:
1. \(4A + 4B = 0\)
2. \(5A - 5B = 1\)

From the first equation, we have \(A + B = 0\) or \(B = -A\). Substituting into the second equation:
\[ 5A - 5(-A) = 1 \implies 10A = 1 \implies A = \frac{1}{10} \]
Thus, \(B = -\frac{1}{10}\).

### Step 5: Rewrite the Integral
Now we can rewrite the integral:
\[ \int \left(\frac{1/10}{4x - 5} - \frac{1/10}{4x + 5}\right) dx \]
This can be simplified to:
\[ \frac{1}{10} \int \frac{dx}{4x - 5} - \frac{1}{10} \int \frac{dx}{4x + 5} \]

### Step 6: Integrate Each Term
The integrals can be solved using the formula \(\int \frac{dx}{ax + b} = \frac{1}{a} \ln |ax + b| + C\):
1. \(\int \frac{dx}{4x - 5} = \frac{1}{4} \ln |4x - 5|\)
2. \(\int \frac{dx}{4x + 5} = \frac{1}{4} \ln |4x + 5|\)

Thus, we have:
\[ \frac{1}{10} \left(\frac{1}{4} \ln |4x - 5| - \frac{1}{4} \ln |4x + 5|\right) + C \]

### Step 7: Combine the Logarithms
Using the properties of logarithms:
\[ \frac{1}{10} \cdot \frac{1}{4} \left(\ln |4x - 5| - \ln |4x + 5|\right) = \frac{1}{40} \ln \left|\frac{4x - 5}{4x + 5}\right| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{dx}{16x^2 - 25} = \frac{1}{40} \ln \left|\frac{4x - 5}{4x + 5}\right| + C \]