Mathematics

# Evaluate $\displaystyle\int_{2}^{4}\dfrac{x}{x^{2}+1}dx$

##### SOLUTION
$\displaystyle\int_{4}^{2}{\dfrac{x}{{x}^{2}+1}dx}$

$=\dfrac{1}{2} \displaystyle\int_{4}^{2}{\dfrac{2x}{{x}^{2}+1}dx}$

Let $t={x}^{2}+1\Rightarrow\,dt=2x\,dx$

$=\dfrac{1}{2} \displaystyle\int{\dfrac{dt}{t}}$

$=\dfrac{1}{2}\left[\log{\left|t\right|}\right]$

$=\dfrac{1}{2}\left[\log{\left|{x}^{2}+1\right|}\right]_{4}^{2}$ where $t={x}^{2}+1$

$=\dfrac{1}{2}\left[\log{\left|{2}^{2}+1\right|}-\log{\left|{4}^{2}+1\right|}\right]$

$=\dfrac{1}{2}\left[\log{\left|4+1\right|}-\log{\left|16+1\right|}\right]$

$=\dfrac{1}{2}\left[\log{\left|5\right|}-\log{\left|17\right|}\right]$

$=\dfrac{1}{2}\log{\left|\dfrac{5}{17}\right|}$

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Subjective Medium Published on 17th 09, 2020
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