$\displaystyle\int_1^e (\ln x)^2\, dx = ?$
- $e - 2$
- $e + 2$
- $e^2 - 1$
- $2e - 1$
Show answer and solution
Correct answer: A. $e - 2$
**IBP:** $u = (\ln x)^2$, $dv = dx$.
$= [x(\ln x)^2]_1^e - \int_1^e 2\ln x\, dx = e \cdot 1 - 0 - 2[x\ln x - x]_1^e = e - 2[(e - e) - (0 - 1)] = e - 2(1) = e - 2$.