`int1/(x^4+3x^2+9)dx`

Answer

To solve the integral \( \int \frac{1}{x^4 + 3x^2 + 9} \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{x^4 + 3x^2 + 9} \, dx \]
To facilitate the integration, we can multiply and divide by 6:
\[ I = \frac{1}{6} \int \frac{6}{x^4 + 3x^2 + 9} \, dx \]

### Step 2: Factor the Denominator
Next, we rewrite the denominator:
\[ x^4 + 3x^2 + 9 = (x^2)^2 + 3(x^2) + 9 \]
Let \( y = x^2 \), then we have:
\[ y^2 + 3y + 9 \]
This can be expressed as:
\[ y^2 + 3y + 9 = (y + \frac{3}{2})^2 + \frac{27}{4} \]
Thus, we can rewrite the integral as:
\[ I = \frac{1}{6} \int \frac{6}{(y + \frac{3}{2})^2 + \frac{27}{4}} \, dy \]

### Step 3: Use Substitution
Now, let:
\[ u = y + \frac{3}{2} \quad \text{so that} \quad dy = du \]
The integral becomes:
\[ I = \frac{1}{6} \int \frac{6}{u^2 + \frac{27}{4}} \, du \]

### Step 4: Simplify the Integral
This integral can be simplified further:
\[ I = \int \frac{1}{u^2 + \left(\frac{3\sqrt{3}}{2}\right)^2} \, du \]
Using the formula:
\[ \int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \]
we find:
\[ I = \frac{1}{\frac{3\sqrt{3}}{2}} \tan^{-1} \left( \frac{u}{\frac{3\sqrt{3}}{2}} \right) + C \]

### Step 5: Substitute Back
Substituting back for \( u \):
\[ I = \frac{2}{3\sqrt{3}} \tan^{-1} \left( \frac{y + \frac{3}{2}}{\frac{3\sqrt{3}}{2}} \right) + C \]
Substituting \( y = x^2 \):
\[ I = \frac{2}{3\sqrt{3}} \tan^{-1} \left( \frac{x^2 + \frac{3}{2}}{\frac{3\sqrt{3}}{2}} \right) + C \]

### Final Answer
Thus, the final answer is:
\[ I = \frac{2}{3\sqrt{3}} \tan^{-1} \left( \frac{2x^2 + 3}{3\sqrt{3}} \right) + C \]