The value of the integral `int_0^(log5)(e^xsqrt(e^x-1))/(e^x+3)dx`

Options

A. `3+2pi`
B. `4-pi`
C. `2+pi`
D. none of these

Explanation

To solve the integral

\[ I = \int_0^{\log 5} \frac{e^x \sqrt{e^x - 1}}{e^x + 3} \, dx, \]

we will use a substitution method. Let's follow the steps:

### Step 1: Substitution
Let \( e^x - 1 = t^2 \). Then, we have:

\[ e^x = t^2 + 1. \]

Differentiating both sides, we get:

\[ \frac{d}{dx}(e^x) \, dx = 2t \, dt \implies e^x \, dx = 2t \, dt. \]

### Step 2: Change of Limits
When \( x = 0 \):

\[ e^0 - 1 = 0 \implies t^2 = 0 \implies t = 0. \]

When \( x = \log 5 \):

\[ e^{\log 5} - 1 = 5 - 1 = 4 \implies t^2 = 4 \implies t = 2. \]

### Step 3: Rewrite the Integral
Now substituting into the integral, we have:

\[ I = \int_0^2 \frac{2t \cdot t}{(t^2 + 1) + 3} \, dt = \int_0^2 \frac{2t^2}{t^2 + 4} \, dt. \]

### Step 4: Simplifying the Integral
We can rewrite the integral as:

\[ I = 2 \int_0^2 \frac{t^2}{t^2 + 4} \, dt. \]

Now, we can simplify this further by adding and subtracting 4 in the denominator:

\[ I = 2 \int_0^2 \left(1 - \frac{4}{t^2 + 4}\right) dt. \]

### Step 5: Break Down the Integral
This gives us:

\[ I = 2 \left( \int_0^2 1 \, dt - \int_0^2 \frac{4}{t^2 + 4} \, dt \right). \]

### Step 6: Evaluate the First Integral
The first integral is straightforward:

\[ \int_0^2 1 \, dt = [t]_0^2 = 2 - 0 = 2. \]

### Step 7: Evaluate the Second Integral
For the second integral, we can use the formula for the integral of \( \frac{1}{a^2 + x^2} \):

\[ \int \frac{4}{t^2 + 4} \, dt = 2 \tan^{-1} \left( \frac{t}{2} \right). \]

Thus,

\[ \int_0^2 \frac{4}{t^2 + 4} \, dt = 2 \left[ \tan^{-1} \left( \frac{t}{2} \right) \right]_0^2 = 2 \left( \tan^{-1}(1) - \tan^{-1}(0) \right) = 2 \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{2}. \]

### Step 8: Combine Results
Now substituting back into our expression for \( I \):

\[ I = 2 \left( 2 - \frac{\pi}{2} \right) = 4 - \pi. \]

### Final Answer
Thus, the value of the integral is:

\[ \boxed{4 - \pi}. \]