`int(1)/(sqrt(2+x-3x^2))dx`

Answer

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

### Step 1: Rewrite the Denominator
We start with the expression under the square root:
\[ \sqrt{2 + x - 3x^2} \]
To make it easier to work with, we can factor out \(-3\) from the quadratic term:
\[ \sqrt{-3\left(x^2 - \frac{1}{3}x - \frac{2}{3}\right)} \]
This gives us:
\[ \sqrt{-3} \sqrt{\frac{1}{3} - x^2 + \frac{1}{3}x} \]
Now, we can multiply and divide by \(\sqrt{3}\):
\[ \sqrt{3} \sqrt{\frac{2}{3} + \frac{1}{3}x - x^2} \]

### Step 2: Completing the Square
Next, we will complete the square for the expression \( -3x^2 + x + 2 \):
\[ -3\left(x^2 - \frac{1}{3}x - \frac{2}{3}\right) \]
The term can be rewritten as:
\[ -3\left(\left(x - \frac{1}{6}\right)^2 - \frac{25}{36}\right) \]
This simplifies to:
\[ -\frac{3}{36}\left(36\left(x - \frac{1}{6}\right)^2 - 25\right) \]
Thus, the expression becomes:
\[ -\frac{1}{12}\left(25 - 36\left(x - \frac{1}{6}\right)^2\right) \]

### Step 3: Substitute and Simplify
Now, substituting back into the integral:
\[ \int \frac{1}{\sqrt{-\frac{1}{12}\left(25 - 36\left(x - \frac{1}{6}\right)^2\right)}} \, dx \]
This can be simplified to:
\[ \int \frac{\sqrt{12}}{\sqrt{25 - 36\left(x - \frac{1}{6}\right)^2}} \, dx \]

### Step 4: Use the Standard Integral Formula
We recognize this as a standard form of the integral:
\[ \int \frac{1}{\sqrt{a^2 - u^2}} \, du = \sin^{-1}\left(\frac{u}{a}\right) + C \]
where \( a = 5 \) and \( u = 6\left(x - \frac{1}{6}\right) \).

### Step 5: Final Integration
Thus, we have:
\[ \frac{1}{\sqrt{3}} \sin^{-1}\left(\frac{6\left(x - \frac{1}{6}\right)}{5}\right) + C \]
Finally, simplifying gives us:
\[ \frac{1}{\sqrt{3}} \sin^{-1}\left(\frac{6x - 1}{5}\right) + C \]

### Final Answer
The final result of the integration is:
\[ \int \frac{1}{\sqrt{2 + x - 3x^2}} \, dx = \frac{1}{\sqrt{3}} \sin^{-1}\left(\frac{6x - 1}{5}\right) + C \]