To solve the integral and find the value of \(\frac{2}{3}(A + B + C + D)\), we follow these steps: Given Integral: \[ \int \frac{\sqrt[4]{x}}{\sqrt{x} + \sqrt[4]{x}} \, dx = \frac{2}{3} \left[ A \sqrt[4]{x^3} + B \sqrt[4]{x^2} + C \sqrt[4]{x} + D \log \left( 1 + \sqrt[4]{x} \right) \right] + K \] Step 1: Simplify the Integral Let’s make a substitution to simplify the integral. Let \( u = \sqrt[4]{x} \). Then, \( x = u^4 \) and \( dx = 4u^3 \, du \). Substituting into the integral: \[ \int \frac{u}{\sqrt{u^4} + u} \cdot 4u^3 \, du = \int \frac{u}{u^2 + u} \cdot 4u^3 \, du = \int \frac{4u^4}{u^2 + u} \, du \] Simplify the integrand: \[ \frac{4u^4}{u^2 + u} = \frac{4u^4}{u(u + 1)} = \frac{4u^3}{u + 1} \] Step 2: Perform Polynomial Division Divide \( 4u^3 \) by \( u + 1 \): \[ 4u^3 = 4u^2(u + 1) - 4u^2 \] \[ -4u^2 = -4u(u + 1) + 4u \] \[ 4u = 4(u + 1) - 4 \] So, \[ \frac{4u^3}{u + 1} = 4u^2 - 4u + 4 - \frac{4}{u + 1} \] Step 3: Integrate Term by Term Integrate each term separately: \[ \int \left( 4u^2 - 4u + 4 - \frac{4}{u + 1} \right) du = \frac{4}{3}u^3 - 2u^2 + 4u - 4 \ln|u + 1| + K \] Step 4: Substitute Back \( u = \sqrt[4]{x} \) \[ \frac{4}{3} (\sqrt[4]{x})^3 - 2 (\sqrt[4]{x})^2 + 4 \sqrt[4]{x} - 4 \ln|1 + \sqrt[4]{x}| + K \] Step 5: Compare with Given Form The given form is: \[ \frac{2}{3} \left[ A \sqrt[4]{x^3} + B \sqrt[4]{x^2} + C \sqrt[4]{x} + D \log \left( 1 + \sqrt[4]{x} \right) \right] + K \] Comparing coefficients: \[ \frac{2}{3} A = \frac{4}{3} \Rightarrow A = 2 \] \[ \frac{2}{3} B = -2 \Rightarrow B = -3 \] \[ \frac{2}{3} C = 4 \Rightarrow C = 6 \] \[ \frac{2}{3} D = -4 \Rightarrow D = -6 \] Step 6: Calculate \( \frac{2}{3}(A + B + C + D) \) \[ A + B + C + D = 2 - 3 + 6 - 6 = -1 \] \[ \frac{2}{3}(A + B + C + D) = \frac{2}{3} \times (-1) = -\frac{2}{3} \] Final Answer: \[ \boxed{-\frac{2}{3}} \]
For the beam and loading shown in the figure, the second derivative of the deflection curve of the beam at the mid-point of AC is given by \( \frac{\alpha M_0}{8EI} \). The value of \( \alpha \) is ........ (rounded off to the nearest integer).
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to:
The integral is given by:
\[ 80 \int_{0}^{\frac{\pi}{4}} \frac{\sin\theta + \cos\theta}{9 + 16 \sin 2\theta} d\theta \]
is equals to?
Arrange the following in increasing order of their pK\(_b\) values.
What is Z in the following set of reactions?
Acetophenone can be prepared from which of the following reactants?