=t,x=2t,dx=2dt ∴I=2‌∫et(sec‌4‌t+4‌tan‌4‌t.sec‌4‌t)dt .......(i) We know, ∫ex(f(x)+f′(x))dx=exf(x)+C .......(ii) and
d
dx
(sec‌4‌x)=4‌sec‌4‌x‌tan‌4‌x From Eqs. (i) and (ii), we get =2[et‌sec‌4‌t+C′]=2et‌sec‌4‌t+2C′ Put the value of t, we get =2ex∕2‌sec‌4.(