Given f(x)=∫0xetf(t)dt+ex⇒f(0)=1
Differentiating with respect to x we get,
f′(x)=exf(x)+ex
f′(x)=ex(f(x)+1)
f(x)+1f′(x)=ex
Integrate both the sides we get,
∫0xf(x)+1f′(x)dx=∫0xexdx
ln(f(x)+1)0x=ex0x
ln(f(x)+1)−ln(f(0)+1)=ex−1
ln(2f(x)+1)=ex−1 {as f(0)=1}
f(x)=2e(ex−1)−1