Mathematics

# Solve $\displaystyle\int\limits_1^3 {\frac{{\log x}}{{{{\left( {x + 1} \right)}^2}}}dx}$

$I=\dfrac{3\log 3}{4}-\log \left( 2 \right)$

##### SOLUTION

Consider the given integral.

$I=\displaystyle\int_{1}^{3}{\left( \dfrac{\log x}{{{\left( x+1 \right)}^{2}}} \right)}dx$

Since, $\displaystyle\int{uvdx=u\displaystyle\int{vdx-\displaystyle\int{\left( \dfrac{d}{dx}\left( u \right)\displaystyle\int{vdx} \right)}}}dx$

Thus,

$I=\displaystyle\int_{1}^{3}{\left( \dfrac{\log x}{{{\left( x+1 \right)}^{2}}} \right)}dx$

$I=\left[ \log x\left( -\dfrac{1}{x+1} \right) \right]_{1}^{3}-\displaystyle\int_{1}^{3}{\dfrac{1}{x}\left( -\dfrac{1}{x+1} \right)}dx$

$I=\left[ -\dfrac{\log x}{x+1} \right]_{1}^{3}+\displaystyle\int_{1}^{3}{\dfrac{1}{x\left( x+1 \right)}}dx$

$I=-\left[ \dfrac{\log x}{x+1} \right]_{1}^{3}+\displaystyle\int_{1}^{3}{\left( \dfrac{1}{x}-\dfrac{1}{x+1} \right)}dx$

$I=-\left[ \dfrac{\log x}{x+1} \right]_{1}^{3}+\left[ \log \left( x \right)-\log \left( x+1 \right) \right]_{1}^{3}$

$I=-\left[ \dfrac{\log 3}{3+1}-\dfrac{\log 1}{1+1} \right]+\left[ \log \left( 3 \right)-\log \left( 3+1 \right)-\left[ \log \left( 1 \right)-\log \left( 1+1 \right) \right] \right]$

$I=-\left[ \dfrac{\log 3}{4}-0 \right]+\left[ \log \left( 3 \right)-\log \left( 4 \right)-\left[ 0-\log \left( 2 \right) \right] \right]$

$I=-\left[ \dfrac{\log 3}{4} \right]+\left[ \log \left( 3 \right)-\log \left( 4 \right)+\log \left( 2 \right) \right]$

$I=-\left[ \dfrac{\log 3}{4} \right]+\left[ \log \left( 3 \right)-2\log \left( 2 \right)+\log \left( 2 \right) \right]$

$I=\log \left( 3 \right)-\log \left( 2 \right)-\dfrac{\log 3}{4}$

$I=\dfrac{3\log 3}{4}-\log \left( 2 \right)$

Hence, this is the answer.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

#### Realted Questions

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$I=\int { \left\{ \log _{ e }( \log _{ e }{ x }) +\cfrac { 1 }{ { \left( \log _{ e }{ x } \right) }^{ 2 } } \right\} } dx$ is equal to
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