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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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1 Verified Answer | Published on 17th 09, 2020