$\int \frac{1}{x} \sec^2(\log x) \, dx = $

$\int \frac{\sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x=$

Integrate the function $\frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}$.

The integral $\int \frac{e^{3 \log_{e} 2x} + 5e^{2 \log_{e} 2x}}{e^{4 \log_{e} x} + 5e^{3 \log_{e} x} - 7e^{2 \log_{e} x}} dx$,where $x > 0$,is equal to ....... (where $c$ is a constant of integration).

For $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$,if $y(x) = \int \frac{\operatorname{cosec} x + \sin x}{\operatorname{cosec} x \sec x + \tan x \sin^2 x} \, dx$ and $\lim_{x \rightarrow (\frac{\pi}{2})^-} y(x) = 0$,then $y\left(\frac{\pi}{4}\right)$ is equal to:

$\int \frac{\sqrt{2} \sin x}{\sin \left(x+\frac{\pi}{4}\right)} d x$ is equal to