`int(4x-3)/(3x^2+2x-5)dx`

Answer

To solve the integral \(\int \frac{4x - 3}{3x^2 + 2x - 5} \, dx\), we can use the method of partial fractions and integration techniques. Here’s a step-by-step solution:

### Step 1: Set up the integral
Let \( I = \int \frac{4x - 3}{3x^2 + 2x - 5} \, dx \).

### Step 2: Rewrite the numerator
We can express the numerator \(4x - 3\) as a linear combination of the derivative of the denominator. We know that the derivative of the denominator \(3x^2 + 2x - 5\) is \(6x + 2\).

We can write:
\[ 4x - 3 = A(6x + 2) + B \]
for some constants \(A\) and \(B\).

### Step 3: Determine coefficients
Expanding the equation gives:
\[ 4x - 3 = (6A)x + (2A + B) \]
Now, we can equate coefficients:
1. For \(x\): \(6A = 4\) \(\Rightarrow A = \frac{2}{3}\)
2. For the constant term: \(2A + B = -3\)

Substituting \(A\) into the second equation:
\[ 2 \cdot \frac{2}{3} + B = -3 \Rightarrow \frac{4}{3} + B = -3 \Rightarrow B = -3 - \frac{4}{3} = -\frac{13}{3} \]

### Step 4: Rewrite the integral
Now we can rewrite the integral:
\[ I = \int \left( \frac{2}{3} \cdot \frac{6x + 2}{3x^2 + 2x - 5} - \frac{13}{3} \cdot \frac{1}{3x^2 + 2x - 5} \right) \, dx \]
This separates into two integrals:
\[ I = \frac{2}{3} \int \frac{6x + 2}{3x^2 + 2x - 5} \, dx - \frac{13}{3} \int \frac{1}{3x^2 + 2x - 5} \, dx \]

### Step 5: Solve the first integral
Using the formula \(\int \frac{f'(x)}{f(x)} \, dx = \ln |f(x)| + C\):
Let \(f(x) = 3x^2 + 2x - 5\), then \(f'(x) = 6x + 2\):
\[ \int \frac{6x + 2}{3x^2 + 2x - 5} \, dx = \ln |3x^2 + 2x - 5| + C \]

Thus,
\[ \frac{2}{3} \int \frac{6x + 2}{3x^2 + 2x - 5} \, dx = \frac{2}{3} \ln |3x^2 + 2x - 5| \]

### Step 6: Solve the second integral
To solve \(\int \frac{1}{3x^2 + 2x - 5} \, dx\), we complete the square:
\[ 3x^2 + 2x - 5 = 3\left(x^2 + \frac{2}{3}x - \frac{5}{3}\right) \]
Completing the square gives:
\[ 3\left((x + \frac{1}{3})^2 - \frac{16}{9}\right) = 3(x + \frac{1}{3})^2 - \frac{16}{3} \]
Thus,
\[ \int \frac{1}{3x^2 + 2x - 5} \, dx = \frac{1}{\sqrt{16/3}} \tan^{-1}\left(\frac{x + \frac{1}{3}}{\sqrt{16/3}}\right) + C = \frac{3}{4} \tan^{-1}\left(\frac{3(x + \frac{1}{3})}{4}\right) \]

### Step 7: Combine results
Putting everything together:
\[ I = \frac{2}{3} \ln |3x^2 + 2x - 5| - \frac{13}{3} \cdot \frac{3}{4} \tan^{-1}\left(\frac{3(x + \frac{1}{3})}{4}\right) + C \]
Simplifying gives:
\[ I = \frac{2}{3} \ln |3x^2 + 2x - 5| - \frac{13}{4} \tan^{-1}\left(\frac{3(x + \frac{1}{3})}{4}\right) + C \]

### Final Answer
\[ \int \frac{4x - 3}{3x^2 + 2x - 5} \, dx = \frac{2}{3} \ln |3x^2 + 2x - 5| - \frac{13}{4} \tan^{-1}\left(\frac{3(x + \frac{1}{3})}{4}\right) + C \]