$$1+x+\int_1^x\left(\text{ln}^2\left(t\right)+2\text{ln}\left(t\right)\right)dt$$
If the above function is f(x) then f'(x) vanishes at which value of x? I am trying this question by taking the integrating part as z and then I am rewriting f(x) =1+x+z. After this I am thinking for applying Newton Leibnitz rule. But the fact is that I am unable to solve any more. Kindly help me out
By Fundamental Theorem of Calculus $f'(x)=1+(\ln x )^{2}+2\ln x$. Hence, $f'(x)=[1+\ln x]^{2}=0$ if and only if $\ln x =-1$ or $x=\frac 1 e$.
First, you derive the function. Then you get: $$1+(\int_1^x(\ln^2(t)+2ln(t))dt)'=1+\ln^2(x)+2ln(x)$$ This happen thanks to the fundamental theorem of calculus. Then, this is a binomial form. So you can simplify to $(1+\ln(x))^2$. Then... ¿For what value you can get to be equal 0? The answer is $x=\frac{1}{e}$.
