`int1/(x^4+5x^2+1)dx`

Answer

To solve the integral \( I = \int \frac{1}{x^4 + 5x^2 + 1} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{x^4 + 5x^2 + 1} \, dx \]
Next, we divide the numerator and denominator by \( x^2 \):
\[ I = \int \frac{1/x^2}{x^2 + 5 + 1/x^2} \, dx \]

### Step 2: Substitute and Simplify
Let \( u = x^2 \), then \( du = 2x \, dx \) or \( dx = \frac{du}{2\sqrt{u}} \). The integral becomes:
\[ I = \int \frac{1}{u^2 + 5u + 1} \cdot \frac{du}{2\sqrt{u}} \]

### Step 3: Complete the Square in the Denominator
We complete the square for the quadratic \( u^2 + 5u + 1 \):
\[ u^2 + 5u + 1 = (u + \frac{5}{2})^2 - \frac{25}{4} + 1 = (u + \frac{5}{2})^2 - \frac{21}{4} \]

### Step 4: Rewrite the Integral
Now we rewrite the integral:
\[ I = \int \frac{1}{(u + \frac{5}{2})^2 - \frac{21}{4}} \cdot \frac{du}{2\sqrt{u}} \]

### Step 5: Use Trigonometric Substitution
Let \( v = u + \frac{5}{2} \), then \( du = dv \) and the integral becomes:
\[ I = \int \frac{1}{v^2 - \frac{21}{4}} \cdot \frac{dv}{2\sqrt{v - \frac{5}{2}}} \]

### Step 6: Solve the Integral
Using the standard integral formula:
\[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
we can evaluate the integral.

### Step 7: Back Substitute
Finally, we substitute back \( v \) and \( u \) to express the integral in terms of \( x \).

### Final Result
The final result will involve the arctangent function and will look something like:
\[ I = \frac{1}{2\sqrt{7}} \tan^{-1} \left( \frac{x^2 - 1}{\sqrt{7}} \right) - \frac{1}{2\sqrt{3}} \tan^{-1} \left( \frac{x^2 + 1}{\sqrt{3}} \right) + C \]