If $f: R^{+} \rightarrow R$ be a differentiable function satisfying $f(x)=x \ln x+x-\int_1^x \frac{f

Question

If $f: R^{+} \rightarrow R$ be a differentiable function satisfying $f(x)=x \ln x+x-\int_1^x \frac{f(t)}{t \cdot \ln t} d t \forall x \in R^{+} \quad$, then the value of $f^{\prime}(e)+f^{\prime \prime}\left(\frac{1}{4}\right)=$

Professor PC Thomas Classes · Answer

$$
\begin{aligned}
& f(x)=x \ln x+x-\int_1^{-} \frac{f(t) a t}{t \ln t} \
& 	herefore f^{\prime}(x)=\frac{x}{x}+\ln x+1-\frac{f(x)}{x \ln x} 	herefore f^{\prime}(x)=\frac{f(x)}{x \ln x}=2+\ln x \
& \Rightarrow \frac{d y}{d x}+\frac{y}{x \ln x}-2+\ln x
\end{aligned}
$$
This is linear differentiable equation.
$$
	ext { ।. } \mathrm{F}=e^{\int \frac{d x}{x \ln x}}=e^{\ln (\ln x)}=\ln x
$$
Solution is
$$
\begin{aligned}
& y \ln x=\int(2+\ln x) \ln x d x \
& 	herefore y \ln x=\int\left(\ln ^2 x+2 \ln xight) d x=x \ln ^2 x+C
\end{aligned}
$$
Put $\mathbf{x}=\mathbf{1}$ get $\mathbf{C}=\mathbf{0}$
$$
\begin{aligned}
& y \ln x=x \ln ^2 x \
& 	herefore \quad y=f(x)=x \ln x \
& 	herefore f^{\prime}(x)=\frac{1}{x} \quad \Rightarrow f^{\prime}(e)+f^{\prime \prime}\left(\frac{1}{4}ight)=(1+\ln e)+4 \quad=2+4=6
\end{aligned}
$$