Evaluate the following integrals
`int (x dx)/(9-16x^(4))`

Answer

To evaluate the integral

\[ I = \int \frac{x \, dx}{9 - 16x^4} \]

we can follow these steps:

### Step 1: Rewrite the integral
First, we can rewrite the denominator in a more manageable form. Notice that \(9\) can be expressed as \(3^2\) and \(16x^4\) can be expressed as \((4x^2)^2\):

\[ I = \int \frac{x \, dx}{3^2 - (4x^2)^2} \]

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

\[ dt = 8x \, dx \quad \Rightarrow \quad x \, dx = \frac{dt}{8} \]

Now substitute \(x \, dx\) and \(t\) into the integral:

\[ I = \int \frac{\frac{dt}{8}}{9 - t^2} = \frac{1}{8} \int \frac{dt}{9 - t^2} \]

### Step 3: Factor the denominator
Next, we can factor the denominator:

\[ 9 - t^2 = (3 - t)(3 + t) \]

### Step 4: Partial Fraction Decomposition
We can express the integrand using partial fractions:

\[ \frac{1}{9 - t^2} = \frac{1}{(3 - t)(3 + t)} = \frac{A}{3 - t} + \frac{B}{3 + t} \]

Multiplying through by the denominator \((3 - t)(3 + t)\) gives:

\[ 1 = A(3 + t) + B(3 - t) \]

Setting \(t = 3\) gives \(1 = 6A \Rightarrow A = \frac{1}{6}\).

Setting \(t = -3\) gives \(1 = 6B \Rightarrow B = \frac{1}{6}\).

Thus, we have:

\[ \frac{1}{9 - t^2} = \frac{1/6}{3 - t} + \frac{1/6}{3 + t} \]

### Step 5: Substitute back into the integral
Now substitute this back into the integral:

\[ I = \frac{1}{8} \left( \frac{1}{6} \int \frac{dt}{3 - t} + \frac{1}{6} \int \frac{dt}{3 + t} \right) \]

### Step 6: Integrate
Integrating each term gives:

\[ I = \frac{1}{48} \left( -\ln |3 - t| + \ln |3 + t| \right) + C \]

Using properties of logarithms, this can be simplified to:

\[ I = \frac{1}{48} \ln \left| \frac{3 + t}{3 - t} \right| + C \]

### Step 7: Substitute back for \(t\)
Recall that \(t = 4x^2\):

\[ I = \frac{1}{48} \ln \left| \frac{3 + 4x^2}{3 - 4x^2} \right| + C \]

### Final Answer
Thus, the final result of the integral is:

\[ I = \frac{1}{48} \ln \left| \frac{3 + 4x^2}{3 - 4x^2} \right| + C \]

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