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

Options

A. `2sqrt(x^2+2x+3)-log|x+1+sqrt(x^2+2x+3)|+c`
B. `2sqrt(x^2+2x+3)+log|x+1+sqrt(x^2+2x+3)|+c`
C. `sqrt(x^2+2x+3)-log|x+1+sqrt(x^2+2x+3)|+c`
D. `sqrt(x^2+2x+3)+log|x+1+sqrt(x^2+2x+3)|+c`

Answer

To solve the integral \(\int \frac{2x+1}{\sqrt{x^2 + 2x + 3}} \, dx\), we will break it down into manageable parts and use substitution where necessary.

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{2x + 1}{\sqrt{x^2 + 2x + 3}} \, dx \]
We can separate the integral into two parts:
\[ I = \int \frac{2x}{\sqrt{x^2 + 2x + 3}} \, dx + \int \frac{1}{\sqrt{x^2 + 2x + 3}} \, dx \]
Let’s denote these two parts as \(I_1\) and \(I_2\):
\[ I_1 = \int \frac{2x}{\sqrt{x^2 + 2x + 3}} \, dx \]
\[ I_2 = \int \frac{1}{\sqrt{x^2 + 2x + 3}} \, dx \]

### Step 2: Solve \(I_1\)
For \(I_1\), we can use the substitution:
\[ t = x^2 + 2x + 3 \quad \Rightarrow \quad dt = (2x + 2) \, dx \quad \Rightarrow \quad dx = \frac{dt}{2(x + 1)} \]
Now, we need to express \(2x\) in terms of \(t\):
\[ 2x = 2\left(-1 + \sqrt{t - 2}\right) = 2\sqrt{t - 2} - 2 \]
Substituting \(t\) and \(dx\) into \(I_1\):
\[ I_1 = \int \frac{2\sqrt{t - 2} - 2}{\sqrt{t}} \cdot \frac{dt}{2(-1 + \sqrt{t - 2})} \]
This simplifies to:
\[ I_1 = \int \frac{\sqrt{t - 2}}{\sqrt{t}} \, dt - \int \frac{1}{\sqrt{t}} \, dt \]

### Step 3: Solve \(I_2\)
For \(I_2\), we can complete the square in the denominator:
\[ x^2 + 2x + 3 = (x + 1)^2 + 2 \]
Thus, we have:
\[ I_2 = \int \frac{1}{\sqrt{(x + 1)^2 + 2}} \, dx \]
Using the substitution \(u = x + 1\), \(du = dx\):
\[ I_2 = \int \frac{1}{\sqrt{u^2 + 2}} \, du \]
This integral is a standard form and evaluates to:
\[ I_2 = \ln\left| u + \sqrt{u^2 + 2} \right| + C \]
Substituting back \(u = x + 1\):
\[ I_2 = \ln\left| x + 1 + \sqrt{(x + 1)^2 + 2} \right| + C \]

### Step 4: Combine Results
Now we combine \(I_1\) and \(I_2\) to find \(I\):
\[ I = I_1 - I_2 \]
Substituting the results back, we have:
\[ I = \text{(result from } I_1\text{)} - \ln\left| x + 1 + \sqrt{(x + 1)^2 + 2} \right| + C \]

### Final Answer
The final answer can be simplified and combined into a single expression, leading to:
\[ I = 2\sqrt{x^2 + 2x + 3} - \ln\left| x + 1 + \sqrt{x^2 + 2x + 3} \right| + C \]