Substitution rule for definite integrals 2

Put $\ln x=t$

Hence, $\frac {1}{x}dx =dt$

The integral reduces to $\int ^{1}_0 tdt = \dfrac {1}{2}$

A simple substitution makes this problem an easy one.

Let $u=ln(x)$ , because its differential $du=\Large \frac{du}{dx}dx =\Large\frac{1}{x}dx$ , which occurs in the integral.

To find the new limits of integration:

$u=ln(e)=1$ (Top limit)

$u=ln(1)=0$ (Bottom limit)

Therefore ${\displaystyle \int_{1}^{e} \frac{\ln(x)}{x}dx}$ becomes ${\displaystyle \int_{0}^{1} \frac{u}{x}dx}$ ,

which is ${\displaystyle \int_{0}^{1} u du}$ ,

which we know is 0.5.