If `f(x) = sqrt(4x^(2)+4x-3)` then `int(x+3)/(f(x))dx` is equal

Options

A. (1/4) f(x) + (2/3) log | 2x + 1+ f(x)| + C
B. (1/4) f(x) + (5/4) log |2x + 1 + f(x)| + C
C. (1/3) f(x) + (2/3) log |(x+3)| + C
D. (2/3) f(x) + (1/3) log |(x+3) + f(x)|

Answer

To solve the integral \(\int \frac{x + 3}{f(x)} \, dx\) where \(f(x) = \sqrt{4x^2 + 4x - 3}\), we will follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \frac{x + 3}{\sqrt{4x^2 + 4x - 3}} \, dx \]

### Step 2: Decompose the numerator
We can express the numerator \(x + 3\) in terms of the derivative of the function inside the square root. We want to find constants \(a\) and \(b\) such that:
\[ x + 3 = a(8x + 4) + b \]
where \(8x + 4\) is the derivative of \(4x^2 + 4x - 3\).

### Step 3: Set up equations
From the equation \(x + 3 = a(8x + 4) + b\), we can compare coefficients:
1. For \(x\): \(8a = 1\) → \(a = \frac{1}{8}\)
2. For the constant term: \(4a + b = 3\)

Substituting \(a = \frac{1}{8}\) into the second equation:
\[ 4 \cdot \frac{1}{8} + b = 3 \implies \frac{1}{2} + b = 3 \implies b = 3 - \frac{1}{2} = \frac{5}{2} \]

### Step 4: Rewrite the integral
Now we can rewrite the integral \(I\) as:
\[ I = \int \frac{1}{8}(8x + 4) \frac{1}{\sqrt{4x^2 + 4x - 3}} \, dx + \frac{5}{2} \int \frac{1}{\sqrt{4x^2 + 4x - 3}} \, dx \]
This gives us:
\[ I = \frac{1}{8} \int \frac{8x + 4}{\sqrt{4x^2 + 4x - 3}} \, dx + \frac{5}{2} \int \frac{1}{\sqrt{4x^2 + 4x - 3}} \, dx \]

### Step 5: Solve the first integral
Let \(t = 4x^2 + 4x - 3\), then \(dt = (8x + 4) \, dx\). The first integral becomes:
\[ \frac{1}{8} \int \frac{dt}{\sqrt{t}} = \frac{1}{8} \cdot 2\sqrt{t} + C_1 = \frac{1}{4} \sqrt{4x^2 + 4x - 3} + C_1 \]

### Step 6: Solve the second integral
Now for the second integral:
\[ \frac{5}{2} \int \frac{1}{\sqrt{4x^2 + 4x - 3}} \, dx \]
We can factor out a 4 from the square root:
\[ = \frac{5}{2} \int \frac{1}{2\sqrt{x^2 + x - \frac{3}{4}}} \, dx = \frac{5}{4} \int \frac{1}{\sqrt{x^2 + x - \frac{3}{4}}} \, dx \]
To solve this integral, we complete the square:
\[ x^2 + x - \frac{3}{4} = \left(x + \frac{1}{2}\right)^2 - 1 \]
Thus, we have:
\[ = \frac{5}{4} \int \frac{1}{\sqrt{\left(x + \frac{1}{2}\right)^2 - 1}} \, dx \]
Using the formula \(\int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln |x + \sqrt{x^2 - a^2}| + C\), we get:
\[ = \frac{5}{4} \ln \left| x + \frac{1}{2} + \sqrt{\left(x + \frac{1}{2}\right)^2 - 1} \right| + C_2 \]

### Step 7: Combine results
Combining both parts, we have:
\[ I = \frac{1}{4} \sqrt{4x^2 + 4x - 3} + \frac{5}{4} \ln \left| 2x + 1 + \sqrt{4x^2 + 4x - 3} \right| + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{x + 3}{\sqrt{4x^2 + 4x - 3}} \, dx = \frac{1}{4} \sqrt{4x^2 + 4x - 3} + \frac{5}{4} \ln \left| 2x + 1 + \sqrt{4x^2 + 4x - 3} \right| + C \]