allery$\int_1^2 \dfrac{\log x}{x^2}\,dx$ equals:
A$(1 - \log 2)/2$
B$\log 2 - 1$
C$1 + \log 2$
D$\log 2/2$
Answer & Solution
Correct answer: A. $(1 - \log 2)/2$
Integration by parts with $u = \log x$, $dv = dx/x^2$ gives $v = -1/x$: $[-\log x/x]_1^2 + \int_1^2 dx/x^2 = (-\log 2/2 - 0) + [-1/x]_1^2 = -\log 2/2 + (1 - 1/2) = (1 - \log 2)/2$.
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