Integrate : x /x4-x2+1

Let I = ∫xx4 - x2 + 1 dx put x2 = t⇒2x dx = dt⇒x dx = dt2So, I = 12∫dtt2 - t + 1=12∫dtt2 - t + 14 - 14 + 1=12∫dtt - 122 +322=12 × 23 tan-1t-1232 + C=13 tan-12t - 13 + C=13 tan-12x2 - 13 + C

Integrate : x /x⁴-x²+1

Let I = \int \frac{x}{x^{4} - x^{2} + 1} d x put x^{2} = t \Rightarrow 2 x d x = d t \Rightarrow x d x = \frac{d t}{2} So, I = \frac{1}{2} \int \frac{d t}{t^{2} - t + 1} = \frac{1}{2} \int \frac{d t}{t^{2} - t + \frac{1}{4} - \frac{1}{4} + 1} = \frac{1}{2} \int \frac{d t}{{(t - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \frac{1}{2} \times \frac{2}{\sqrt{3}} \tan^{- 1} [\frac{t - \frac{1}{2}}{\frac{\sqrt{3}}{2}}] + C = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 t - 1}{\sqrt{3}}) + C = \frac{1}{\sqrt{3}} \tan^{- 1} (\frac{2 x^{2} - 1}{\sqrt{3}}) + C

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