Let f(x) be a function satisfying f′(x)=f(x) and
f(0)=2. Then ∫f(x)3+4f(x)dx
(1/4)log3+8ex+C
(1/2)log3+5ex+C
(1/4)log3+4e2x+C
none of these
First show that f(x)=2ex. Now,
I=∫2ex3+8exdx=14log3+8ex+C