Evaluate the following
`int (2x+9)/(x^2+9x+30) dx.`

Answer

To evaluate the integral

\[ \int \frac{2x + 9}{x^2 + 9x + 30} \, dx, \]

we can follow these steps:

### Step 1: Simplify the Integral
We can separate the integral into two parts:

\[ \int \frac{2x}{x^2 + 9x + 30} \, dx + \int \frac{9}{x^2 + 9x + 30} \, dx. \]

### Step 2: Evaluate the First Integral
For the first integral, we can use substitution. Let

\[ t = x^2 + 9x + 30. \]

Then, differentiate \(t\) with respect to \(x\):

\[ dt = (2x + 9) \, dx. \]

This means that

\[ dx = \frac{dt}{2x + 9}. \]

Substituting \(t\) into the integral gives:

\[ \int \frac{2x + 9}{t} \cdot \frac{dt}{2x + 9} = \int \frac{dt}{t} = \ln |t| + C = \ln |x^2 + 9x + 30| + C. \]

### Step 3: Evaluate the Second Integral
For the second integral, we need to evaluate

\[ \int \frac{9}{x^2 + 9x + 30} \, dx. \]

To do this, we can complete the square for the quadratic in the denominator:

\[ x^2 + 9x + 30 = (x + \frac{9}{2})^2 + \frac{120}{4} - \frac{81}{4} = (x + \frac{9}{2})^2 + \frac{39}{4}. \]

Now, we can rewrite the integral:

\[ \int \frac{9}{(x + \frac{9}{2})^2 + \frac{39}{4}} \, dx. \]

Using the formula for the integral of the form \(\int \frac{a}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C\), we have:

Here \(a^2 = \frac{39}{4}\), so \(a = \frac{\sqrt{39}}{2}\).

Thus, the integral becomes:

\[ 9 \cdot \frac{2}{\sqrt{39}} \tan^{-1} \left(\frac{2(x + \frac{9}{2})}{\sqrt{39}}\right) + C = \frac{18}{\sqrt{39}} \tan^{-1} \left(\frac{2(x + \frac{9}{2})}{\sqrt{39}}\right) + C. \]

### Step 4: Combine the Results
Combining both parts, we have:

\[ \int \frac{2x + 9}{x^2 + 9x + 30} \, dx = \ln |x^2 + 9x + 30| + \frac{18}{\sqrt{39}} \tan^{-1} \left(\frac{2(x + \frac{9}{2})}{\sqrt{39}}\right) + C. \]

### Final Answer
Thus, the final result is:

\[ \int \frac{2x + 9}{x^2 + 9x + 30} \, dx = \ln |x^2 + 9x + 30| + \frac{18}{\sqrt{39}} \tan^{-1} \left(\frac{2(x + \frac{9}{2})}{\sqrt{39}}\right) + C. \]

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