`int(1)/(sqrt(4x^(2)-x+4))dx`

Answer

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

### Step 1: Rewrite the integral
We start by rewriting the integral:
\[ I = \int \frac{1}{\sqrt{4x^2 - x + 4}} \, dx \]

### Step 2: Factor out the constant from the square root
We can factor out 4 from the expression under the square root:
\[ I = \int \frac{1}{\sqrt{4\left(x^2 - \frac{x}{4} + 1\right)}} \, dx = \frac{1}{2} \int \frac{1}{\sqrt{x^2 - \frac{x}{4} + 1}} \, dx \]

### Step 3: Complete the square
Next, we complete the square for the expression \( x^2 - \frac{x}{4} + 1 \):
\[ x^2 - \frac{x}{4} + 1 = \left(x - \frac{1}{8}\right)^2 + \left(1 - \frac{1}{64}\right) = \left(x - \frac{1}{8}\right)^2 + \frac{63}{64} \]
Thus, we can rewrite the integral as:
\[ I = \frac{1}{2} \int \frac{1}{\sqrt{\left(x - \frac{1}{8}\right)^2 + \left(\frac{\sqrt{63}}{8}\right)^2}} \, dx \]

### Step 4: Use the standard integral formula
We can now use the standard integral formula:
\[ \int \frac{1}{\sqrt{u^2 + a^2}} \, du = \ln \left| u + \sqrt{u^2 + a^2} \right| + C \]
In our case, \( u = x - \frac{1}{8} \) and \( a = \frac{\sqrt{63}}{8} \).

### Step 5: Substitute into the formula
Applying the formula, we get:
\[ I = \frac{1}{2} \ln \left| x - \frac{1}{8} + \frac{\sqrt{63}}{8} \right| + \frac{1}{2} \ln \left| \sqrt{\left(x - \frac{1}{8}\right)^2 + \left(\frac{\sqrt{63}}{8}\right)^2} \right| + C \]

### Step 6: Simplify the expression
We can combine the logarithmic terms:
\[ I = \frac{1}{2} \ln \left| 8x - 1 + \sqrt{(8x - 1)^2 + 63} \right| + C \]

### Final Answer
Thus, the final result for the integral is:
\[ I = \frac{1}{2} \ln \left| 8x - 1 + \sqrt{(8x - 1)^2 + 63} \right| + C \]