Question

Evaluate

${\int }_1^2\frac{\ell nx}{x^2}dx$

Answer 1

We have,

${\int }_1^2\frac{\ln x}{x^2}dx$

$={{[-\frac{\ln x}{x}]}_1}^2+{\int }_1^2\left[\left(\frac{d}{dx}\ln x\right)x^{-2}dx\right]dx$

$=\frac{-1}{2}\ln 2+\ln 1+{\int }_1^2\left[\frac{1}{x}{x}^{-2}dx\right]dx$

$=\frac{-1}{2}\ln 2+\ln 1+{\int }_1^2\left[x^{-3}dx\right]dx$

$=\frac{-1}{2}\ln 2+\ln 1+{\int }_1^2\left[\frac{x^{-2}}{-2}\right]dx$

$=\frac{-1}{2}\ln 2+\ln 1+\frac{1}{-2}{\int }_1^2{x}^{-2}dx$

$=\frac{-1}{2}\ln 2+\ln 1-\frac{1}{2}{{\left[\frac{x^{-1}}{-1}\right]}_1}^2+C$

$=\frac{-1}{2}\ln 2+\ln 1+\frac{1}{2}{{\left[x^{-1}\right]}_1}^2+C$

$=\frac{-1}{2}\ln 2+\ln 1+\frac{1}{2}{{[2^{-1}-1^{-1}]}_1}^2+C$

$=\frac{-1}{2}\ln 2+\ln 1+\frac{1}{2}[\frac{1}{2}-1]+C$

$=\frac{-1}{2}\ln 2+\ln 1+\frac{1}{2}(-\frac{1}{2})+C$

$=\frac{-1}{2}\ln 2+0-\frac{1}{4}+C\therefore \ln 1=0$

$=\frac{-1}{2}\ln 2-\frac{1}{4}+C$

Hence, this is the answer.