Solve the following differential equation: $\left(x^2+1\right) \frac{dy}{dx} + 4xy = \frac{1}{x^2+1}$

$A$ function $y = f(x)$ satisfies the differential equation $f(x) \sin 2x - \cos x + (1 + \sin^2 x) f'(x) = 0$ where $f(0) = 0$. Then the value of $f(\frac{\pi}{6})$ is equal to

The integrating factor of the differential equation $x \frac{dy}{dx} + y \log x = x e^x \cdot x^{-1/2} \log x$ for $x > 0$ is:

Let us consider a curve $y=f(x)$ passing through the point $(-2, 2)$ and the slope of the tangent to the curve at any point $(x, f(x))$ is given by $f(x)+x f'(x)=x^2$. Then:

Let $f$ be a real-valued differentiable function on $\mathbb{R}$ (the set of all real numbers) such that $f(1)=1$. If the $y$-intercept of the tangent at any point $P(x, y)$ on the curve $y=f(x)$ is equal to the cube of the abscissa of $P$,then the value of $f(-3)$ is equal to

The integrating factor of the differential equation $3 x \log _{e} x \frac{d y}{d x}+y=2 \log _{e} x$ is given by