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Find:

\int_{a}^{b}\frac{\ log x}{x}dx

 

 

 

 
 
 
 
 

Answers (1)

I=\int_{a}^{b}\frac{\ log x}{x}dx

Put logx=t \: \: \therefore \frac{1}{x}dx=dt

x=a, t=loga

x=b, t=logb

I=\int_{\log a}^{\log b}tdt

I=\left [ \frac{t^2}{2} \right ]^{logb}_{loga}

I=\frac{1}{2}\left [ \left ( logb \right )^2-\left ( loga \right )^2 \right ]

I=\frac{1}{2}\left [ \left ( logb+loga \right )\left ( logb-loga \right ) \right ]

I=\frac{1}{2}\log{(ab)}\log \frac{b}{a}

Posted by

Ravindra Pindel

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