The integral ( x 8 - x 2 ) d x ( x 12 + 3 x 6 + 1 ) tan - 1 ( x 3 + 1 x 3 ) is equal to: [2024]

Question

The integral $\int \frac{(x^{8} - x^{2}) d x}{(x^{12} + 3 x^{6} + 1) \tan^{- 1} (x^{3} + \frac{1}{x^{3}})}$ is equal to: [2024]

Smart Achievers · Accepted Answer

(3)

Let $I = \int \frac{(x^{8} - x^{2}) d x}{(x^{12} + 3 x^{6} + 1) \tan^{- 1} (x^{3} + \frac{1}{x^{3}})}$

Put $\tan^{- 1} (x^{3} + \frac{1}{x^{3}}) = u$

$\Rightarrow \frac{1}{1 + {(x^{3} + \frac{1}{x^{3}})}^{2}} \cdot 3 (x^{2} - \frac{1}{x^{4}}) d x = d u$

$\Rightarrow (\frac{x^{6}}{x^{12} + 3 x^{6} + 1} \times \frac{x^{6} - 1}{x^{4}}) d x = \frac{d u}{3}$

$\Rightarrow I = \frac{1}{3} \int \frac{d u}{u} = \frac{\ln | u |}{3} = \ln$ $(| \tan^{- 1} {(x^{3} + \frac{1}{x^{3}}) |)}^{\frac{1}{3}} + C$

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