By using the properties of definite integrals,evaluate the integral $\int_{0}^{4}|x-1| d x$.

$\int_{-1}^1 \frac{\log (1+x)}{1+x^2} d x = \int_0^1 \frac{\log (1+x)}{1+x^2} d x + \int_0^1 f(x) d x$, then $f(x) =$

The value of the integral $\int_0^\infty \frac{\log_e(x)}{x^2+4} dx$ is:

$\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1-\sin x \cdot \cos x} d x$ is equal to

The value of the integral $\int_{-\pi}^{\pi} (\cos px - \sin qx)^2 dx$,where $p$ and $q$ are integers,is equal to:

Let $f(x) = 7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$ for all $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the correct expression$(s)$ is(are):
$(A) \int_0^{\pi/4} x f(x) dx = \frac{1}{12}$
$(B) \int_0^{\pi/4} f(x) dx = 0$
$(C) \int_0^{\pi/4} x f(x) dx = \frac{1}{6}$
$(D) \int_0^{\pi/4} f(x) dx = 1$