Mathematics

Evaluate $$\displaystyle \int \frac{x^4 + 1}{x^6 + 1} dx $$


SOLUTION
$$I = \displaystyle \int \frac{x^4 + 1}{x^6 + 1} dx = \displaystyle \int \frac{(x^2 + 1)^2 - 2x^2}{(x^2 + 1) (x^4 - x^2 + 1)} dx$$
  $$= \displaystyle \int \frac{(1 + 1/x^2)dx}{(x^2 + 1/x^2 - 1)} - 2 \displaystyle \int \frac{x^2 dx}{(x^3)^2 + 1}$$
  $$ = \displaystyle \int \frac{(1 + 1/x^2) dx}{(x - 1/x)^2 + 1} - 2 \displaystyle \int \frac{x^2 dx}{(x^3)^2 + 1}$$
 In first integral put $$x - 1/x = t $$ and in second integral put $$x^3 = u$$
           $$= \displaystyle \int \frac{dt}{t^2 + 1} - \frac{3}{2} \displaystyle \int \frac{du}{u^2 + 1}$$
           $$= \tan^{-1}(t) - \frac{2}{3} \tan^{-1} (u) + C$$
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Subjective Medium Published on 17th 09, 2020
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