Mathematics

# $\displaystyle \int\frac{\log(x+1)-\log x}{x(x+1)}dx=$

$-\displaystyle \frac{1}{2}[\log (1+\displaystyle \frac{1}{x})]^{2}+c$

##### SOLUTION
$\displaystyle I=\int\dfrac{\log(x+1)-\log x}{x(x+1)}dx$

$\displaystyle =\int \dfrac { \log (\dfrac { x+1 }{ x } ) }{ x(x+1) } dx$

Put $\displaystyle \log (\dfrac { x+1 }{ x } )=t$
$\Rightarrow \displaystyle \dfrac { -1 }{ x(x+1) } dx=dt$

So, $I=-\int tdt$
$\displaystyle \Rightarrow I=-\dfrac { { t }^{ 2 } }{ 2 } +C$
$\displaystyle I=-\dfrac { { [\log (\dfrac { x+1 }{ x } )] }^{ 2 } }{ 2 } +C$
$I=-\displaystyle \dfrac{1}{2}[\log (1+\displaystyle \dfrac{1}{x})]^{2}+c$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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