Let $\int (2 x^{6} + 15 x^{4} + 2 x^{2} + 3) \cos 2 x d x = f (x) . \sin 2 x + g (x) . \cos 2 x + k$ where f(x) and g(x) are polynomial functions of x and k is constant of integration, then $\int \frac{g (x)}{f (x)} d x =$

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$\frac{1}{2} \tan^{- 1} (x^{6} + x^{2} + 1) + c$

$\frac{1}{2} \tan^{- 1} (x^{6} + x^{4} + 1) + c$

$\frac{1}{2} \log (x^{6} + x^{2} + 1) + c$

$\frac{1}{2} \log (x^{6} + x^{4} + 1) + c$

answer is A.

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Detailed Solution

$\int (2 x^{6} + 15 x^{4} + 2 x^{2} + 3) \cos 2 x d x = \int (\underset{I}{\underset{⏟}{2 x^{6} + 2 x^{2}}}) \underset{I I}{\underset{⏟}{\cos 2 x d x}} + \int \underset{I I}{\underset{⏟}{(15 x^{4} + 3)}} \underset{I}{\underset{⏟}{\cos 2 x d x}} = (2 x^{6} + 2 x^{2}) \frac{\sin 2 x}{2} - \int (12 x^{5} + 4 x) \frac{\sin 2 x}{2} d x + \cos 2 x (\frac{15 x^{5}}{5} + 3 x) + \int 2 \sin 2 x . (\frac{15 x^{5}}{5} + 3 x) d x = (x^{6} + x^{2}) \sin 2 x - \int (6 x^{5} + 2 x) \sin 2 x d x + (3 x^{5} + 3 x) \cos 2 x + \int 6 \sin 2 x (x^{5} + x) d x = (x^{6} + x^{2}) \sin 2 x + 3 (x^{5} + x) \cos 2 x + 4 \int x \sin 2 x d x = (x^{6} + x^{2}) \sin 2 x + 3 (x^{5} + x) \cos 2 x + 4 [x (\frac{- \cos 2 x}{2}) + \int \frac{\cos 2 x}{2} d x] = \underset{f (x)}{\underset{⏟}{(x^{6} + x^{2} + 1)}} \sin 2 x + \underset{g (x)}{\underset{⏟}{(3 x^{5} + x)}} \cos 2 x + k (i) \int \frac{3 x^{5} + x}{x^{6} + x^{2} + 1} d x = \frac{1}{2} l n (x^{6} + x^{4} + 1) + C$