etf(t)dt+ex . . . (i) Since, f(x) is differentiable function, differentiate Eq. (i) f′(x)=exf(x)+ex[Using Newton Leibnitz theorem ] f′(x)=ex(f(x)+1)⇒‌
f′(x)
f(x)+1
=ex Integrating it, ∫‌
f′(x)
f(x)+1
dx=∫exdx+C Let f(x)+1=u, then f′(x)dx=du ∫‌
du
u
=ex+C⇒log‌u=ex+C log(f(x)+1)=ex+C‌‌‌‌[∵u=f(x)+1]... (ii) Now f(x)=
x
∫
0
etf(t)dt+ex⇒f(0)=e0=1 Put x=0, in Eq. (ii), we get log(2)=e∘+C⇒C=log(2)−1 From Eq. (ii), we get log(f(x)+1)‌‌=ex+log‌2−1 f(x)+1‌‌=eex+log‌2−1=eex⋅elog‌2⋅e−1 f(x)+1‌‌=2eex⋅e−1=2eex−1 ∴‌‌f(x)‌‌=2eex−1−1