The integral _1^e( xe^2x-e^xx)_ex,dx is equal to :

Question

The integral $\displaystyle\int_1^e\left(\frac x{e^{2x}}-\frac{e^x}x\right)\log_ex\,dx$ is equal to :

Accepted Answer

Let $\dfrac{x}{e^x}=t$; then $x\ln x-x=\ln t$, so on differentiating, $\ln x\,dx=\dfrac{dt}t$. When $x=1$, $t=\dfrac1e$; when $x=e$, $t=1$.

$$I=\int_{1/e}^1\left(t^2-\frac1t\right)\cdot\frac{dt}t=\int_{1/e}^1\left(t-\frac1{t^2}\right)dt=\left[\frac{t^2}2+\frac1t\right]_{1/e}^1$$

$$=\left(\frac12+1\right)-\left(\frac1{2e^2}+e\right)=\frac32-e-\frac1{2e^2}$$

Solutions · 1