$$ \int_1^4\left(x+\sqrt{x}+\frac{1}{x}\right) d x-\int_1^{2 \log 2} d x= $$

Let $I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{2-\cos 2 x}\left(\frac{\beta}{\pi}+\log \left(\frac{4+\sin x}{4-\sin x}\right)\right) d x$. Given that $\int \frac{d x}{1+k x^2}=\frac{1}{\sqrt{k}} \tan ^{-1}(\sqrt{k} x)+c, \tan ^{-1}(0)=0$ and $\tan ^{-1}(\sqrt{3})=\pi / 3$. Then, $3 I^2=$

If $f(x)=\frac{1}{x^3} \int_5^x\left(2 u^2-u f^{\prime}(u) d u\right.$, then $f^{\prime}(5)=$

Assertion (A) $\int\limits_{-a}^a f(x) d x=\int_0^a(f(x)+f(-x)) d x$

Reason (R) $\int\limits_a^b f(x) d x=\int_{g(a)}^{g(b)} f(g(u)) g^{\prime}(u) d u$

The correct option among the following is