Originally Posted by

**skeeter** $\displaystyle \int_1^e \dfrac{1}{x} \, dx = \int_1^e x \cdot \dfrac{1}{x^2} \, dx$

$u = x \implies du = dx$

$dv = \dfrac{1}{x^2} \, dx \implies v = -\dfrac{1}{x}$

$\displaystyle \int_1^e \dfrac{1}{x} \, dx = \left[x \cdot \left(-\dfrac{1}{x}\right) \right]_1^e - \int_1^e -\dfrac{1}{x} \, dx$

$\displaystyle \int_1^e \dfrac{1}{x} \, dx = \bigg[-1 \bigg]_1^e + \int_1^e \dfrac{1}{x} \, dx$

$\displaystyle \int_1^e \dfrac{1}{x} \, dx - \int_1^e \dfrac{1}{x} \, dx = \bigg[-1 \bigg]_1^e = [-1 - (-1)] = 0$