Mathematics

# Evaluate the following integral:$\displaystyle \int { \cfrac { 1 }{ 2{ x }^{ 2 }-x-1 } } dx\quad$

##### SOLUTION
$\displaystyle\int{\dfrac{dx}{2{x}^{2}-x-1}}$

$=\displaystyle\int{\dfrac{dx}{2\left({x}^{2}-\dfrac{1}{2}x-\dfrac{1}{2}\right)}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dx}{\left({x}^{2}-\dfrac{1}{2}x-\dfrac{1}{2}\right)}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dx}{\left({x}^{2}-2\times\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}-\dfrac{1}{2}\right)}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dx}{\left({\left(x-\dfrac{1}{4}\right)}^{2}-\dfrac{1}{16}-\dfrac{8}{16}\right)}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dx}{\left({\left(x-\dfrac{1}{4}\right)}^{2}-\dfrac{9}{16}\right)}}$

$=\dfrac{1}{2}\displaystyle\int{\dfrac{dx}{\left({\left(x-\dfrac{1}{4}\right)}^{2}-{\left(\dfrac{3}{4}\right)}^{2}\right)}}$

We know that $\displaystyle\int{\dfrac{dx}{{x}^{2}-{a}^{2}}}=\dfrac{1}{2a}\log{\left|\dfrac{x-a}{x+a}\right|}+c$

Replace $x\rightarrow\,x-\dfrac{1}{4}$ and $a\rightarrow\,\dfrac{3}{4}$ we get

$=\dfrac{1}{2}\times\dfrac{1}{2\times\dfrac{3}{4}}\log{\left|\dfrac{x-\dfrac{1}{4}-\dfrac{3}{4}}{x-\dfrac{1}{4}+\dfrac{3}{4}}\right|}+c$

$=\dfrac{1}{2}\times\dfrac{2}{3}\log{\left|\dfrac{x-\dfrac{4}{4}}{x+\dfrac{2}{4}}\right|}+c$

$=\dfrac{1}{2}\times\dfrac{2}{3}\log{\left|\dfrac{x-1}{x+\dfrac{1}{2}}\right|}+c$

$=\dfrac{1}{3}\log{\left|\dfrac{2x-2}{2x+1}\right|}+c$

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Subjective Medium Published on 17th 09, 2020
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