Mathematics

# Evaluate $\displaystyle\int_{1}^{2}\dfrac{1}{x(1+\log x)^{2}}dx$

##### SOLUTION
$\displaystyle\int_{1}^{2}{\dfrac{1}{x{\left(1+\log{x}\right)}^{2}}dx}$

Let $t=1+\log{x}\Rightarrow\,dt=\dfrac{1}{x}dx$

$=\displaystyle\int{\dfrac{dt}{{t}^{2}}}$

$=\displaystyle\int{{t}^{-2}dt}$

$=\left[\dfrac{{t}^{-2+1}}{-2+1}\right]_{1}^{2}$

$=\left[-\dfrac{1}{t}\right]$

$=\left[-\dfrac{1}{1+\log{x}}\right]_{1}^{2}$ where $t=1+\log{x}$

$=-\left[\dfrac{1}{1+\log{x}}\right]_{1}^{2}$

$=-\left[\dfrac{1}{1+\log{2}}-\dfrac{1}{1+\log{1}}\right]$

$=-\left[\dfrac{1}{1+\log{2}}-\dfrac{1}{1+0}\right]$

$=-\left[\dfrac{1}{1+\log{2}}-1\right]$

$=-\left[\dfrac{1-1-\log{2}}{1+\log{2}}\right]$

$=-\left[\dfrac{-\log{2}}{1+\log{2}}\right]$

$=\dfrac{\log{2}}{1+\log{2}}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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