If $\int\left[\left(\frac{x}{e}\right)^x+\left(\frac{e}{x}\right)^x\right] \ln x d x=A\left(\frac{x}

Question

If $\int\left[\left(\frac{x}{e}\right)^x+\left(\frac{e}{x}\right)^x\right] \ln x d x=A\left(\frac{x}{e}\right)^x+B\left(\frac{e}{x}\right)^x+C$, then the value of $\mathrm{A}+\mathrm{B}$ is

Professor PC Thomas Classes · Answer

$\begin{aligned} & \left.\int\left(\frac{x}{e}ight)^x+\left(\frac{e}{x}ight)^xight] \ln x d x \ & 	ext { Put } \quad\left(\frac{x}{e}ight)^x=t \quad 	ext { Or } \quad x \ln \left(\frac{x}{e}ight)=\ln t \ & 	herefore \quad\left(x \cdot \frac{1}{x / e} \frac{1}{e}+\ln \left(\frac{x}{e}ight)ight) d x=\frac{1}{t} d t \quad 	herefore \quad(1+\ln x-\ln e) d x=\frac{1}{t} d t \ & 	herefore \quad(\ln e+\ln x-\ln e) d x=\frac{1}{t} d t \ & 	herefore \quad(\ln x) d x=\frac{1}{t} d t \ & 	ext { Or } \quad I=\int\left(t+\frac{1}{t}ight) \frac{1}{t} d t=\int 1 \cdot d t+\int \frac{1}{t^2} d t=t-\frac{1}{t}+C\end{aligned}$