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

Options

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

Answer

To solve the integral \( \int \frac{3x + 2}{\sqrt{2x^2 + 2x + 1}} \, dx \), we will follow these steps:

### Step 1: Simplify the Denominator
First, we simplify the expression under the square root:
\[ 2x^2 + 2x + 1 = 2(x^2 + x + \frac{1}{2}) \]
This can be rewritten as:
\[ 2\left(x^2 + x + \frac{1}{4} + \frac{1}{4}\right) = 2\left((x + \frac{1}{2})^2 + \frac{1}{4}\right) \]
Thus, we have:
\[ \sqrt{2x^2 + 2x + 1} = \sqrt{2}\sqrt{(x + \frac{1}{2})^2 + \frac{1}{4}} \]

### Step 2: Substitution
Let:
\[ y = 2x^2 + 2x + 1 \]
Then, differentiating \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 4x + 2 \quad \Rightarrow \quad dy = (4x + 2) \, dx \]
From this, we can express \( dx \) in terms of \( dy \):
\[ dx = \frac{dy}{4x + 2} \]

### Step 3: Rewrite the Integral
Now, substituting \( y \) into the integral:
\[ \int \frac{3x + 2}{\sqrt{y}} \cdot \frac{dy}{4x + 2} \]
We can express \( 3x + 2 \) in terms of \( y \):
\[ 3x + 2 = \frac{3}{4}(4x + 2) - \frac{1}{4} \]
Thus, the integral becomes:
\[ \int \left(\frac{3}{4} - \frac{1}{4(4x + 2)}\right) \frac{dy}{\sqrt{y}} \]

### Step 4: Separate the Integral
This gives us two separate integrals:
\[ \frac{3}{4} \int \frac{dy}{\sqrt{y}} - \frac{1}{4} \int \frac{dy}{(4x + 2)\sqrt{y}} \]

### Step 5: Solve the Integrals
The first integral:
\[ \frac{3}{4} \int \frac{dy}{\sqrt{y}} = \frac{3}{4} \cdot 2\sqrt{y} = \frac{3}{2} \sqrt{y} \]
The second integral can be solved using the substitution \( u = 4x + 2 \), which will require additional steps.

### Step 6: Back Substitute
After solving the integrals, we substitute back \( y = 2x^2 + 2x + 1 \) to express the result in terms of \( x \).

### Final Result
Combining all parts, we will have:
\[ \int \frac{3x + 2}{\sqrt{2x^2 + 2x + 1}} \, dx = \frac{3}{2} \sqrt{2x^2 + 2x + 1} - \text{(other terms)} + C \]