Question

Evaluate: ${\int }_1^2\log xdx$

• A. $2\log 2-1$
• B. $\log 2-1$
• C. $2\log 2+1$
• D. $\log 2-2$

Answer 1

The correct option is A $2\log 2-1$

${\int }_1^2\log xdx$

$={\int }_1^{2}1.\log xdx$

$={[\log x\int 1.dx-\int (\frac{d}{dx}\log x\int 1.dx)]}_1^2$

$={[x\log x-x]}_1^2$

$=(2\log 2-2)-(1\log 1-1)$

$=2\log 2-1$