`int1/(x^2(x^4+1)^(3/4))dx`

Answer

To solve the integral \( I = \int \frac{1}{x^2 (x^4 + 1)^{3/4}} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{x^2 (x^4 + 1)^{3/4}} \, dx \]

### Step 2: Factor out \( x^4 \)
Notice that we can factor \( x^4 \) from \( x^4 + 1 \):
\[ I = \int \frac{1}{x^2 \left( x^4 \left(1 + \frac{1}{x^4}\right) \right)^{3/4}} \, dx \]
This simplifies to:
\[ I = \int \frac{1}{x^2 \cdot x^{3} \cdot \left(1 + \frac{1}{x^4}\right)^{3/4}} \, dx \]
\[ = \int \frac{1}{x^5 \left(1 + \frac{1}{x^4}\right)^{3/4}} \, dx \]

### Step 3: Substitute \( t = 1 + \frac{1}{x^4} \)
Let \( t = 1 + \frac{1}{x^4} \). Then, we differentiate:
\[ dt = -\frac{4}{x^5} \, dx \quad \Rightarrow \quad dx = -\frac{x^5}{4} \, dt \]
Substituting \( x^5 = \frac{1}{(t - 1)^{1/4}} \) into the integral gives:
\[ I = \int \frac{-\frac{x^5}{4} \, dt}{x^5 t^{3/4}} = -\frac{1}{4} \int t^{-3/4} \, dt \]

### 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
Substituting back for \( t \):
\[ I = -\left(1 + \frac{1}{x^4}\right)^{1/4} + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = -\left(1 + \frac{1}{x^4}\right)^{1/4} + C \]