Evaluate ∫(from a to b) (log x)/x dx.

Steps to Solve

1. Identify the integral to evaluate

We need to evaluate the definite integral: $$ \int_{a}^{b} \frac{\log x}{x} , dx $$

2. Use a substitution method

Let ( u = \log x ). Then, the derivative of ( u ) gives: $$ \frac{du}{dx} = \frac{1}{x} \implies dx = x , du = e^u , du $$ Substituting ( x = e^u ) makes the integral become: $$ \int \frac{u}{e^u} , du $$

3. Solve the integral using integration by parts

Using integration by parts, where:

• Let ( v = u ) and ( dw = e^{-u} , du )
• Then ( dv = du ) and ( w = -e^{-u} )

The formula for integration by parts is: $$ \int v , dw = vw - \int w , dv $$

Applying this, we have: $$ \int u e^{-u} , du = -u e^{-u} - \int -e^{-u} , du $$ So, $$ = -u e^{-u} + e^{-u} + C $$

4. Substituting back to ( x )

Now, revert ( u ) back to ( \log x ): $$ = -(\log x) \frac{1}{x} + \frac{1}{x} + C $$

5. Evaluate the definite integral from ( a ) to ( b )

Now, evaluate at the limits ( a ) and ( b ): $$ I = \left[ -\frac{\log x}{x} + \frac{1}{x} \right]_{a}^{b} $$

Calculating gives: $$ I = \left( -\frac{\log b}{b} + \frac{1}{b} \right) - \left( -\frac{\log a}{a} + \frac{1}{a} \right) $$