Mathematics

$$ \int{ \dfrac {1} {{ x }^{ 2 } - {x} + {1}} dx}$$


ANSWER

$$= \dfrac {2}{\sqrt3}\tan^{-1} (\dfrac{2x-1}{\sqrt3})+C$$


SOLUTION
$$ \int{ \dfrac {1} {{ x }^{ 2 } - {x} + {1}} dx}$$
$$=  \int{ \dfrac {1} {{ (x- \dfrac{1}{2}})^{ 2 } + \dfrac {3}{4}} dx}$$
$$= \int{ \dfrac {1} {{ (x- \dfrac{1}{2}})^{ 2 } + (\dfrac {\sqrt3}{2})^2} dx}$$
$$= \dfrac {1}{\dfrac {\sqrt{3}}{2}}{\tan^{-1}}(\dfrac{x-\dfrac 12}{\dfrac{\sqrt3}2})+C $$
$$= \dfrac {2}{\sqrt3}\tan^{-1} (\dfrac{2x-1}{\sqrt3})+C$$
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Single Correct Medium Published on 17th 09, 2020
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