Evaluate the following integrals
`int (x^(3)dx)/(16x^(8)-25)`

Answer

To evaluate the integral

\[ I = \int \frac{x^3 \, dx}{16x^8 - 25}, \]

we can follow these steps:

### Step 1: Rewrite the Denominator
We can rewrite the denominator \(16x^8 - 25\) as a difference of squares:

\[ 16x^8 - 25 = (4x^4)^2 - 5^2. \]

### Step 2: Substitution
Let \(t = 4x^4\). Then, we differentiate \(t\) with respect to \(x\):

\[ dt = 16x^3 \, dx \quad \Rightarrow \quad dx = \frac{dt}{16x^3}. \]

### Step 3: Substitute in the Integral
Now, substitute \(t\) and \(dx\) back into the integral:

\[ I = \int \frac{x^3}{(4x^4)^2 - 5^2} \cdot \frac{dt}{16x^3} = \frac{1}{16} \int \frac{dt}{t^2 - 25}. \]

### Step 4: Simplify the Integral
Now, we can simplify the integral:

\[ I = \frac{1}{16} \int \frac{dt}{t^2 - 5^2}. \]

### Step 5: Use Partial Fraction Decomposition
Using the identity for the difference of squares, we can express the integral as:

\[ \frac{1}{t^2 - 25} = \frac{1}{(t - 5)(t + 5)}. \]

We can use partial fractions:

\[ \frac{1}{(t - 5)(t + 5)} = \frac{A}{t - 5} + \frac{B}{t + 5}. \]

Multiplying through by the denominator:

\[ 1 = A(t + 5) + B(t - 5). \]

Setting \(t = 5\) gives \(A = \frac{1}{10}\) and setting \(t = -5\) gives \(B = -\frac{1}{10}\).

### Step 6: Rewrite the Integral
Thus, we can rewrite the integral as:

\[ I = \frac{1}{16} \int \left( \frac{1/10}{t - 5} - \frac{1/10}{t + 5} \right) dt. \]

### Step 7: Integrate
Now we can integrate:

\[ I = \frac{1}{16} \cdot \frac{1}{10} \left( \ln |t - 5| - \ln |t + 5| \right) + C. \]

### Step 8: Substitute Back
Substituting back \(t = 4x^4\):

\[ I = \frac{1}{160} \left( \ln |4x^4 - 5| - \ln |4x^4 + 5| \right) + C. \]

### Step 9: Combine Logarithms
Using properties of logarithms:

\[ I = \frac{1}{160} \ln \left| \frac{4x^4 - 5}{4x^4 + 5} \right| + C. \]

Thus, the final answer is:

\[ \int \frac{x^3 \, dx}{16x^8 - 25} = \frac{1}{160} \ln \left| \frac{4x^4 - 5}{4x^4 + 5} \right| + C. \]