Mathematics

$$\int {\dfrac{{\log \left( {x + 1} \right) - \log x}}{{x\left( {x + 1} \right)}}} \,\,dx$$ equals


SOLUTION
$$\int { \dfrac { log\left( x+1 \right) -logx }{ x\left( x+1 \right)  } dx } $$
Put $$log\left( x+1 \right) -logx=4\\ \left( \dfrac { 1 }{ x+1 } -\dfrac { 1 }{ x }  \right) dx=dt\\ \dfrac { -1 }{ x\left( x+1 \right)  } dx-dt\\ \dfrac { dx }{ x\left( x+1 \right)  } =-dt$$
$$\int { -tdt } \\ =\dfrac { -{ t }^{ 2 } }{ 2 } +c\\ =\dfrac { -\left[ log\left( x+1 \right) -logx \right] ^{ 2 } }{ 2 } +c$$

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Subjective Medium Published on 17th 09, 2020
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