The integral ∫ ( x ^ 8 - x ^ 2 ) dx ( x ^ 12 + 3 x ^ 6 + 1) ^ - 1 (...

Let, $I = ∫ \dfrac{{x}^{8} - {x}^{2}}{\left({x}^{12} + 3{x}^{6} + 1\right)\tan^{- 1}\left({x}^{3} + \dfrac{1}{{x}^{3}}\right)}dx$

Putting, $\tan^{- 1}\left({x}^{3} + \dfrac{1}{{x}^{3}}\right) = t$

$⇒ \dfrac{1}{1 + {\left({x}^{3} + \dfrac{1}{{x}^{3}}\right)}^{2}} \cdot \left(3{x}^{2} - \dfrac{3}{{x}^{4}}\right)dx = dt$

$⇒ \dfrac{1}{1 + \left({x}^{6} + \dfrac{1}{{x}^{6}} + 2\right)} \cdot \left(3{x}^{2} - \dfrac{3}{{x}^{4}}\right)dx = dt$

$⇒ \dfrac{{x}^{6}}{{x}^{12} + 3{x}^{6} + 1} \cdot \dfrac{3{x}^{6} - 3}{{x}^{4}}dx = dt$

$⇒ \dfrac{\left({x}^{8} - {x}^{2}\right)}{{x}^{12} + 3{x}^{6} + 1}dx = \dfrac{dt}{3}$

$⇒ I = \dfrac{1}{3} ∫ \dfrac{1}{t}dt$

$⇒ I = \dfrac{1}{3}log | t | + C$

⇒ I = \dfrac{1}{3}\log \left|\tan^{- 1}\left({x}^{3} + \dfrac{1}{{x}^{3}}\right)\right| + C$</p><p>$⇒ I = \log \left|\tan^{- 1}\left({x}^{3} + \dfrac{1}{{x}^{3}}\right)\right|\dfrac{1}{3} + C$</p><p>Hence optin \left(A\right)$ is correct