Question

Evaluate the integral $\int_{0}^{1} \frac{1}{\sqrt{3+x} + \sqrt{1+x}} \, dx$ Given that the integral can be expressed in the form $a + b\sqrt{2} + c\sqrt{3}$, where $a, b, c$ are rational numbers, find the value of $2a + 3b - 4c$.

• A. $4$
• B. $10$
• C. $7$
• D. $8$

Answer 1

Answer: (D)

$8$

Answer 2

Given:

$\int_0^1 \frac{1}{\sqrt[3]{x} + \sqrt[3]{x} + \sqrt[3]{1 + x}} \, dx$

Step 1: Rationalizing the Denominator

Rationalize the denominator:
$\int \frac{\sqrt[3]{x + \sqrt{x}} - \sqrt[3]{x - \sqrt{x}}}{\left(\sqrt[3]{x + \sqrt{x}} + \sqrt[3]{x - \sqrt{x}}\right)} \, dx = \int \frac{\sqrt[3]{x + \sqrt{x}}}{2} \, dx$

Step 2: Separating the Integral

Separate the integral:

$\frac{1}{2} \left( \int \sqrt[3]{1 + \sqrt{x}} \, dx - \int \sqrt[3]{1 - \sqrt{x}} \, dx \right)$

Step 3: Evaluating the Integrals

1. For $\int \sqrt[3]{1 + \sqrt{x}} \, dx$:
$\int \sqrt[3]{1 + \sqrt{x}} \, dx = \frac{3}{2} \cdot \frac{3}{4} \cdot 2 + \frac{2}{5} \Rightarrow \frac{3}{2} \left( 2 + \sqrt[3]{3} - 2^{3/2} \right) = \frac{3}{2} (3 - 3\sqrt{3})$

2. For $\int \sqrt[3]{1 - \sqrt{x}} \, dx$:

$\int \sqrt[3]{1 - \sqrt{x}} \, dx = \frac{3}{2} (3 - \sqrt{3}) = \frac{3}{2} (2\sqrt{5} - 1)$

Step 4: Combining the Results

Combine the results:

$\frac{3}{2} (3 + \sqrt{3}) - \frac{3}{2} (3\sqrt{3} - 1) = a + b \sqrt{2 + \sqrt{3}}$

From this, we find:
$a = 3, \quad b = -\frac{2}{3}, \quad c = -1$
Calculate:
$2a + 3b - c = 2\frac{4}{3} + 3\frac{-4}{3} - 4(-1)=8$