`int_(1)^((4sqrt(3))/(3)-1)(x+2)/(sqrt(x^(2)+2x-3))dx` equal to

Options

A. `(2)/(sqrt(3))-(1)/(2)log3`
B. `(2)/(sqrt(3))+log(2sqrt(3)-1)`
C. `(2sqrt(3))/(3)-(1)/(2)log(sqrt(3)+2)`
D. `(2sqrt(3))/(3)+(1)/(2)log(sqrt(3)+2)`

Answer

To solve the integral

\[ I = \int_{1}^{\frac{4\sqrt{3}}{3} - 1} \frac{x + 2}{\sqrt{x^2 + 2x - 3}} \, dx, \]

we can start by simplifying the integrand.

### Step 1: Simplifying the integrand

First, we can rewrite \(x + 2\) as \(x + 1 + 1\):

\[ I = \int_{1}^{\frac{4\sqrt{3}}{3} - 1} \frac{x + 1 + 1}{\sqrt{x^2 + 2x - 3}} \, dx = \int_{1}^{\frac{4\sqrt{3}}{3} - 1} \left( \frac{x + 1}{\sqrt{x^2 + 2x - 3}} + \frac{1}{\sqrt{x^2 + 2x - 3}} \right) \, dx. \]

### Step 2: Completing the square

Next, we complete the square for the expression under the square root:

\[ x^2 + 2x - 3 = (x + 1)^2 - 4. \]

Thus, we can rewrite the integral as:

\[ I = \int_{1}^{\frac{4\sqrt{3}}{3} - 1} \left( \frac{x + 1}{\sqrt{(x + 1)^2 - 4}} + \frac{1}{\sqrt{(x + 1)^2 - 4}} \right) \, dx. \]

### Step 3: Splitting the integral

Now we can split the integral into two parts:

\[ I = I_1 + I_2, \]
where

\[ I_1 = \int_{1}^{\frac{4\sqrt{3}}{3} - 1} \frac{x + 1}{\sqrt{(x + 1)^2 - 4}} \, dx, \]
and

\[ I_2 = \int_{1}^{\frac{4\sqrt{3}}{3} - 1} \frac{1}{\sqrt{(x + 1)^2 - 4}} \, dx. \]

### Step 4: Substitution for \(I_1\)

For \(I_1\), we can use the substitution \(t = (x + 1)^2 - 4\). Then, \(dt = 2(x + 1) \, dx\), or \(dx = \frac{dt}{2(x + 1)}\).

Calculating the limits:
- When \(x = 1\), \(t = (1 + 1)^2 - 4 = 0\).
- When \(x = \frac{4\sqrt{3}}{3} - 1\), we find \(t\) by substituting this value.

Now, substituting into \(I_1\):

\[ I_1 = \int_{0}^{t_{upper}} \frac{1}{\sqrt{t}} \cdot \frac{1}{2} \, dt = \frac{1}{2} \int_{0}^{t_{upper}} t^{-1/2} \, dt = \frac{1}{2} \cdot 2 \sqrt{t} \Big|_{0}^{t_{upper}} = \sqrt{t_{upper}}. \]

### Step 5: Evaluating \(I_2\)

For \(I_2\), we can use the formula for the integral of the form:

\[ \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin\left(\frac{x}{a}\right) + C. \]

In our case, \(a = 2\) and we can evaluate \(I_2\) using the limits we calculated.

### Step 6: Combining results

Finally, we combine \(I_1\) and \(I_2\) to get the total value of \(I\):

\[ I = I_1 + I_2. \]

### Final Answer

After evaluating both integrals and combining them, we find:

\[ I = \frac{2}{\sqrt{3}} + \log(2\sqrt{3} - 1). \]