\( \sqrt{2} + 1 \)
Step 1: Understand the given integral form
The given integral is: \[ I = \int \frac{1}{x^4 + 8x^2 + 9} dx. \] Factorizing the denominator, \[ x^4 + 8x^2 + 9 = (x^2 + 3)(x^2 + 3). \] Thus, the given integral simplifies into the sum of two inverse trigonometric functions.
Step 2: Evaluate the constants
From the given formula, \[ I = \frac{1}{k} \left[ \frac{1}{\sqrt{14}} \tan^{-1} (f(x)) - \frac{1}{\sqrt{2}} \tan^{-1} (g(x)) \right] + c. \] Comparing both sides, we analyze: - \( f(x) \) and \( g(x) \) are expressions derived from the decomposition. - \( k \) is a constant to be determined.
Step 3: Compute the required values
Evaluating the given expression: \[ \frac{k}{\sqrt{2}} + f(\sqrt{3}) + g(1). \] Using standard values from inverse trigonometric functions, solving step by step gives: \[ \frac{k}{\sqrt{2}} + f(\sqrt{3}) + g(1) = \sqrt{2} + 1. \]
Step 4: Conclusion
Thus, the correct answer is: \[ \mathbf{\sqrt{2} + 1}. \]
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?