`int (dx)/(x^(2)(x^(4)+1)^(3/4))is`

Options

A. `-((1)/(x^(4))+1)^(3//4)+c`
B. `(x^(-4)+1)^(3//4)+c`
C. `-((1)/x^(4)+1)^(3//4)+c`
D. `-((1)/x^(4)+1)^(1//4)+c`

Answer

To solve the integral

\[ I = \int \frac{dx}{x^2 (x^4 + 1)^{3/4}}, \]

we will follow these steps:

### Step 1: Simplify the Denominator
We can factor out \(x^4\) from the denominator:

\[ I = \int \frac{dx}{x^2 (x^4 + 1)^{3/4}} = \int \frac{dx}{x^2 \left[ x^4 \left(1 + \frac{1}{x^4}\right)\right]^{3/4}}. \]

This simplifies to:

\[ I = \int \frac{dx}{x^2 x^{3} \left(1 + \frac{1}{x^4}\right)^{3/4}} = \int \frac{dx}{x^{5} \left(1 + \frac{1}{x^4}\right)^{3/4}}. \]

### Step 2: Substitute
Let

\[ t = 1 + \frac{1}{x^4}. \]

Then, differentiating both sides gives:

\[ \frac{dt}{dx} = -\frac{4}{x^5} \implies dx = -\frac{x^5}{4} dt. \]

### Step 3: Substitute into the Integral
Now substituting \(dx\) in the integral:

\[ I = \int \frac{-\frac{x^5}{4} dt}{x^{5} \left(t\right)^{3/4}} = -\frac{1}{4} \int \frac{dt}{t^{3/4}}. \]

### Step 4: Integrate
Now we can integrate:

\[ -\frac{1}{4} \int t^{-3/4} dt = -\frac{1}{4} \cdot \frac{t^{1/4}}{1/4} + C = -t^{1/4} + C. \]

### Step 5: Substitute Back
Now substitute \(t\) back in terms of \(x\):

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

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

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