Mathematics

# The value of $\int _{ }^{ }{ \cfrac { \log { x } }{ { \left( x+1 \right) }^{ 2 } } } dx$ is

$\cfrac { -\log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C$

##### SOLUTION
$\int \dfrac{lnx}{(x+1)^{2}}$
$lnx\int \dfrac{dx}{(x+1)^{2}}-\int \dfrac{1}{x}\int \dfrac{dx}{(x+1)^{2}}$
$lnx\dfrac{(x+1)^{-2+1}}{-2+1}-\int \dfrac{1}{x}\dfrac{(x+1)^{-2+1}}{-2+1}dx$
$-\dfrac{lnx}{(x+1)}+\int \dfrac{1}{x}\dfrac{1}{(x+1)}dx$
$-\dfrac{lnx}{(x+1)}+\int \dfrac{(x+1)-(x)}{x(x+1)}$
$-\dfrac{lnx}{(x+1)}+\int \dfrac{1}{x}-\dfrac{1}{1+x}dx$
$-\dfrac{lnx}{(1+x)}+lnx-ln(1+x)+C$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 128

#### Realted Questions

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$\int _{ 0 }^{ 1 }{ \cot ^{ 1 }{ \left( 1-x+{ x }^{ 2 } \right) dx } }$ equals-
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1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
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Q4 Subjective Medium
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