$\int e^x \left( \frac{2+\sin 2x}{1+\cos 2x} \right) dx$ is equal to

The solution of $\frac{dy}{dx} = e^x(\sin x + \cos x)$ is

Evaluate the integral: $\int \frac{x e^{2x}}{(1+2x)^2} dx = $ (where $C$ is a constant of integration.)

$\int e^x \left( \log x + \frac{1}{x} - \frac{1}{x} + \frac{1}{x^2} \right) dx$ is not the original,the question is $\int e^x \left( \log x + \frac{1}{x} \right) dx$ or similar. Given the options,the integral is $\int e^x \left( \log x + \frac{1}{x} \right) dx$. Let us evaluate $\int e^x \left( \log x + \frac{1}{x} \right) dx$.

$\int \left( \frac{2 - \sin 2x}{1 - \cos 2x} \right) e^x \, dx$ is equal to

If $\int e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx = f(x) + \text{constant}$, then $f(x)$ is equal to