Mathematics

Integrate: $$\int \dfrac{x}{x^4-x^2+1}\ dx$$


SOLUTION
Solution:-
$$\int{\cfrac{x}{{x}^{4} - {x}^{2} + 1} dx}$$
Let $${x}^{2} = t \Rightarrow x \; dx = \cfrac{dt}{2}$$
$$\therefore \int{\cfrac{x}{{x}^{4} - {x}^{2} + 1} dx} = \cfrac{1}{2} \int{\cfrac{dt}{{t}^{2} - t + 1}}$$
$$\Rightarrow = \cfrac{1}{2} \int{\cfrac{dt}{{\left( t - \cfrac{1}{2} \right)}^{2} - \cfrac{1}{4} + 1}}$$
$$\Rightarrow = \cfrac{1}{2} \int{\cfrac{dt}{{\left( t - \cfrac{1}{2} \right)}^{2} + {\left( \cfrac{\sqrt{3}}{2} \right)}^{2}}}$$
$$\Rightarrow = \cfrac{1}{2} \left[ \dfrac{1}{\left( \cfrac{\sqrt{3}}{2} \right)} \tan^{-1}{\dfrac{\left( t - \dfrac{1}{2} \right)}{\left( \cfrac{\sqrt{3}}{2} \right)}} \right] + C$$
$$\Rightarrow = \cfrac{\sqrt{3}}{4} \left( \tan^{-1}{\left( \cfrac{2t - 1}{\sqrt{3}} \right)} \right) + C$$
As $$t = {x}^{2}$$
$$\therefore \int{\cfrac{x}{{x}^{4} - {x}^{2} + 1} dx} = \cfrac{\sqrt{3}}{4} \left( \tan^{-1}{\left( \cfrac{2{x}^{2} - 1}{\sqrt{3}} \right)} \right) + C$$
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