Question

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Answer 1

Topic :-

Indefinite Integration

To Solve :-

$\displaystyle \int \dfrac{x^2-\sqrt{3}x+1}{x^4-x^2+1}\:dx$

Solution :-

$\displaystyle \int \dfrac{x^2-\sqrt{3}x+1}{x^4-x^2+1}\:dx$

Factoring denominator,

$\displaystyle \int \dfrac{x^2-\sqrt{3}x+1}{(x^2+\sqrt{3}x+1)(x^2-\sqrt{3}x+1)}\:dx$

$\displaystyle \int \dfrac{1}{x^2+\sqrt{3}x+1}\:dx$

Completing the square,

$\displaystyle \int \dfrac{1}{\left(x+\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}}\:dx$

$\displaystyle \int \dfrac{1}{\dfrac{(2x+\sqrt{3})^2+1}{4}}\:dx$

$\displaystyle \int \dfrac{4}{(2x+\sqrt{3})^2+1}\:dx$

Substitute 2x + √3 = t,

$2\:dx=dt$

$4\:dx=2\:dt$

$\displaystyle \int \dfrac{2}{t^2+1}\:dt$

$\displaystyle 2\int \dfrac{1}{t^2+1}\:dt$

$2\tan^{-1}t+C$

$\left(\because \displaystyle \int\dfrac{1}{1+z^2}\:dz=\tan^{-1}z+C \right)$

Put back value of 't',

$2\tan^{-1}(2x+\sqrt{3})+C$

$C\:is\:constant\:of\:integration.$

Answer :-

$\underline{\boxed{\displaystyle \int \dfrac{x^2-\sqrt{3}x+1}{x^4-x^2+1}\:dx=2\tan^{-1}(2x+\sqrt{3})+C}}$