Question

$\int [{\left(\frac{x}{e}\right)}^x+{\left(\frac{e}{x}\right)}^x]ln\left(x\right),dx$

• A. ${\left(\frac{x}{e}\right)}^x-{\left(\frac{e}{x}\right)}^x+C$
• B. ${\left(\frac{x}{e}\right)}^x+{\left(\frac{e}{x}\right)}^x+C$
• C. ${\left(-\frac{x}{e}\right)}^x+{\left(\frac{e}{x}\right)}^x+C$
• D. ${\left(-\frac{x}{e}\right)}^x-{\left(\frac{e}{x}\right)}^x+C$

Answer 1

The correct option is A ${\left(\frac{x}{e}\right)}^x-{\left(\frac{e}{x}\right)}^x+C$

Let $I=\int [{\left(\frac{x}{e}\right)}^x+{\left(\frac{e}{x}\right)}^x]lnxdx=\int (1+\frac{1}{{\left(\frac{x}{e}\right)}^{2x}})lnx{\left(\frac{x}{e}\right)}^{x}dx$

Substitute ${\left(\frac{x}{e}\right)}^x=t\Rightarrow differentiatewrttot\Rightarrow lnx{\left(\frac{x}{e}\right)}^{x}dx=dt$

$I=\int (1+\frac{1}{t^2})dt=t-\frac{1}{t}+C={\left(\frac{x}{e}\right)}^x-{\left(\frac{e}{x}\right)}^x+C$