The value of $\int {\frac{1}{{{{(x - 5)}^2}}}\,dx} $ is

If $\int \cos x \cdot \cos 2 x \cdot \cos 5 x \, dx = A \sin 2 x + B \sin 4 x + C \sin 6 x + D \sin 8 x + k$ (where $k$ is the arbitrary constant of integration),then $\frac{1}{B} + \frac{1}{C} = $

$\int \tan ^{-1}\left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) d x=$

If $\int \frac{e^x-1}{e^x+1} dx = f(x) + c$, then $f(x)$ is equal to

$\int \tan^4 x \, dx = $

$\int \frac{\sin^3(x) + \cos^3(x)}{\sin^2(x) \cdot \cos^2(x)} \, dx = $