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:
\(2a + 3b - c = 2\frac{4}{3} + 3\frac{-4}{3} - 4(-1)=8\)
Find the area of the region (in square units) enclosed by the curves: \[ y^2 = 8(x+2), \quad y^2 = 4(1-x) \] and the Y-axis.
Evaluate the integral: \[ I = \int_{\frac{1}{\sqrt[5]{32}}}^{\frac{1}{\sqrt[5]{31}}} \frac{1}{\sqrt[5]{x^{30} + x^{25}}} dx. \]
Evaluate the integral: \[ I = \int_{-3}^{3} |2 - x| dx. \]
Evaluate the integral: \[ I = \int_{-\pi}^{\pi} \frac{x \sin^3 x}{4 - \cos^2 x} dx. \]
If \[ \int \frac{3}{2\cos 3x \sqrt{2} \sin 2x} dx = \frac{3}{2} (\tan x)^{\beta} + \frac{3}{10} (\tan x)^4 + C \] then \( A = \) ?
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32