Mathematics

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


ANSWER

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


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
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