If $\int e^x \left( \frac{x^2-8x+19}{(x-1)^5} \right) dx = \frac{e^x(lx+m)}{(x-1)^4} + C$, then $4l+m=$

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 $(R)$: $\int_a^b e^x (f(x) + f'(x)) dx = e^b f(b) - e^a f(a)$

The value of $I = \int \frac{(x-1) e^x}{(x+1)^3} \,dx$ is

$\int \left( {1 + x - \frac{1}{x}} \right){e^{x + \frac{1}{x}}}\,dx = $

$\int e^x \left( \frac{1+\sin x}{1+\cos x} \right) dx =$

If $\int e^{\sin x}(1+\sec x \tan x) d x=e^{\sin x} f(x)+c$,then in $0 \leq x \leq 2 \pi$,the number of solutions of $f(x)=1$ is