$\int_1^2 \dfrac{\log x}{x^2}\,dx$ 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 $v = -1/x$: $[-\log x/x]_1^2 + \int_1^2 dx/x^2 = (-\log 2/2 - 0) + [-1/x]_1^2 = -\log 2/2 + (1 - 1/2) = (1 - \log 2)/2$.
Related questions
Using definite integration as the limit of a sum, $\int_1^2 (2x + 5)\,dx$ equals:$\int_0^{\pi/2} \cos^2 x\,dx$ equals:$\int_0^{\pi/2} qrt{1 - \cos 4x}\,dx$ equals:Using the King's property, evaluate $\int_0^{\pi/2} \dfrac{dx}{1 + qrt[3]{\tan x}}$:$\int_{-1}^{1} |5x - 3|\,dx$ equals:Evaluate $\int_2^3 7^x\,dx$:$\int_0^4 (x - x^2)\,dx$ equals:$\int_0^1 e^x\,dx$ equals: