The value of `int_0^1(sqrt(x)dx)/((x+1)(3x+1)(x+3))` is

Options

A. `pi/8-(3pi)/16`
B. `pi/8+(3pi)/16`
C. `pi/4-(3pi)/16`
D. 0

Answer

To solve the integral

\[ I = \int_0^1 \frac{\sqrt{x}}{(x+1)(3x+1)(x+3)} \, dx, \]

we will use a substitution method. Let's proceed step by step.

### Step 1: Substitution
Let \( t = \sqrt{x} \). Then, we have:

\[ x = t^2 \quad \text{and} \quad dx = 2t \, dt. \]

When \( x = 0 \), \( t = 0 \) and when \( x = 1 \), \( t = 1 \). Thus, the limits of integration change from \( 0 \) to \( 1 \).

### Step 2: Rewrite the Integral
Substituting \( x \) and \( dx \) into the integral, we get:

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

### Step 3: Simplify the Numerator
We can rewrite the numerator \( 2t^2 \) as:

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

Thus, we can express the integral as:

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

### Step 4: Split the Integral
Now, we can split the integral into two parts:

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

### Step 5: Simplify Each Integral
The first integral simplifies to:

\[ \int_0^1 \frac{1}{(t^2 + 3)(t^2 + 1)} \, dt, \]

and the second integral simplifies to:

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

### Step 6: Evaluate the Integrals
Now we need to evaluate these integrals.

1. For the first integral, we can use partial fractions.
2. For the second integral, we can also use partial fractions.

### Step 7: Combine the Results
After evaluating both integrals, we combine the results to find the value of \( I \).

### Final Result
After performing the calculations and applying limits, we find:

\[ I = \frac{\pi}{8} - \frac{\sqrt{3} \pi}{16}. \]