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

$\int e^{-2 x}\left(\tan 2 x-2 \sec ^2 2 x \tan 2 x\right) d x=$

$\int_0^1 \frac{e^x(x - 1)}{(x + 1)^3} \, dx = $

Assertion $(A)$: $\int_2^e \left(\frac{1}{\log_e x} - \frac{1}{(\log_e x)^2}\right) dx = e - 2 \log_2 e$
Reason $(R)$: $\int_a^b e^x (f(x) + f'(x)) dx = e^b f(b) - e^a f(a)$

$\int e^{x} \sec x(1+\tan x) d x$ equals

The value of $\int \frac{e^{\tan ^{-1}(x)}}{1+x^2} \left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$,for $x>0$ is