Mathematics

$$\displaystyle \int {\dfrac{{\log \left( {x + 1} \right)}}{{\left( {x + 1} \right)}}\ dx} $$ equal to


SOLUTION

Consider the given integral.

$$I=\int{\dfrac{{{\log }_{e}}\left( 1+x \right)}{\left( 1+x \right)}}dx$$

 

Let $$t={{\log }_{e}}\left( 1+x \right)$$

$$dt=\dfrac{dx}{1+x}$$

 

Therefore,

$$ I=\int{t}dt $$

$$ I=\dfrac{{{t}^{2}}}{2}+C $$

 

On putting the value of $$t$$, we get

$$I=\dfrac{{{\left( {{\log }_{e}}\left( 1+x \right) \right)}^{2}}}{2}+C$$

 

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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