Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De bepaalde integraal $https://latex.codecogs.com/png.image?\dpi{110}\int_{1}^{2}\frac{\ln x}{x^2}\; dx$ is gelijk aan	A.
B. ln 2
C. 1 − ln 2
D. ln 2
E. (1 − ln 2)

$https://latex.codecogs.com/png.image?\dpi{110}\int_{1}^{2}\frac{\ln x}{x^2}\; dx=$	A.
B. ln 2
C. − ln 2
D. ln 2
E. ln 2
F. (1 − ln 2)

1^ste manier :
We berekenen eerst de onbepaalde integraal door middel van partiële integratie : $\int\frac{\ln}{x^2}\,dx=\int\ln x\;d\left(-\frac1x\right)=\ln x\cdot\left(-\frac1x\right)-\int\left(-\frac1x\right)d\ln x=-\frac{\ln x}{x}+\int\frac{1}{x^2}\,dx=-\frac{\ln x}{x}-\frac1x+C$
$\int_{1}^{2}\frac{\ln x}{x^2}\,dx=\left[-\frac{\ln x}{x}-\frac1x\right]^2_1=-\frac{\ln2}{2}-\frac12-(-0-1)=-\frac{\ln2}{2}+\frac12=\frac12(1-\ln2)$
2^de manier :
$\textbf{D}\,\frac{\ln x}{x}=\frac{\frac1x.x-(\ln x).1}{x^2}=\frac{1-\ln x}{x^2}=\frac{1}{x^2}-\frac{\ln x}{x^2}$
Daaruit volgt dat $\large\frac{\ln x}{x^2}=\frac{1}{x^2}-D\frac{\ln x}{x}$ zodat
$\int_{1}^{2}\frac{\ln x}{x^2}\,dx=\int_{1}^{2}\frac{1}{x^2}\,dx\;-\int_{1}^{2}D\frac{\ln x}{x}dx=\left[-\frac1x\right]_1^2-\left[\frac{\ln x}{x}\right]_1^2=\left(-\frac12+1\right)-\left(\frac{\ln2}{2}\right)-0=\frac12-\frac{\ln2}{2}=\frac{1-\ln2}{2}\approx 0,1534$