`int(x)/(2x+3)dx=`

Options

A. `(2)/(x)+(4)/(3)logx+c`
B. `(x)/(2)+(x^(2))/(6)+c`
C. `(x)/(2)-(3)/(4)log(2x+3)+c`
D. `(x)/(2)+(4)/(3)log(2x+3)+c`

Answer

To solve the integral \(\int \frac{x}{2x+3} \, dx\), we can use a method of substitution and algebraic manipulation. Let's break it down step by step.

### Step 1: Rewrite the integrand
We start by rewriting the integrand \(\frac{x}{2x+3}\) in a more manageable form. We can express \(x\) in terms of \(2x + 3\):

\[ x = \frac{1}{2}(2x + 3) - \frac{3}{2} \]

This allows us to rewrite the integral:

\[ \int \frac{x}{2x+3} \, dx = \int \frac{\frac{1}{2}(2x + 3) - \frac{3}{2}}{2x + 3} \, dx \]

### Step 2: Simplify the integrand
Now we can simplify the integrand:

\[ \int \left( \frac{1}{2} - \frac{3/2}{2x + 3} \right) \, dx \]

This separates into two integrals:

\[ \int \frac{1}{2} \, dx - \frac{3}{2} \int \frac{1}{2x + 3} \, dx \]

### Step 3: Integrate each term
Now we can integrate each term separately.

1. The first integral:

\[ \int \frac{1}{2} \, dx = \frac{1}{2} x \]

2. The second integral requires a simple substitution. The integral of \(\frac{1}{u}\) is \(\ln |u|\):

\[ -\frac{3}{2} \int \frac{1}{2x + 3} \, dx = -\frac{3}{2} \cdot \frac{1}{2} \ln |2x + 3| = -\frac{3}{4} \ln |2x + 3| \]

### Step 4: Combine the results
Now we combine the results of the two integrals:

\[ \int \frac{x}{2x + 3} \, dx = \frac{1}{2} x - \frac{3}{4} \ln |2x + 3| + C \]

where \(C\) is the constant of integration.

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

\[ \int \frac{x}{2x + 3} \, dx = \frac{x}{2} - \frac{3}{4} \ln |2x + 3| + C \]