The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
The minimum value of the function \( f(x) = x^4 - 4x - 5 \), where \( x \in \mathbb{R} \), is:
The critical points of the function \( f(x) = (x-3)^3(x+2)^2 \) are:
\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]
Then \( f'(-4) \) is equal to:If \( f'(x) = 4x\cos^2(x) \sin\left(\frac{x}{4}\right) \), then \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:
\[ \int \frac{4x \cos \left( \sqrt{4x^2 + 7} \right)}{\sqrt{4x^2 + 7}} \, dx \]
\[ \int \left( \frac{\log_e t}{1+t} + \frac{\log_e t}{t(1+t)} \right) dt \]