The integral $\int \frac{3 x^{13} + 2 x^{11}}{{(2 x^{4} + 3 x^{2} + 1)}^{4}} d x$ is equal to: (where C is a constant of integration)

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$\frac{x^{12}}{6 {(2 x^{4} + 3 x^{2} + 1)}^{3}} + C$

$\frac{x^{12}}{{(2 x^{4} + 3 x^{2} + 1)}^{3}} + C$

$\frac{x^{4}}{{(2 x^{4} + 3 x^{2} + 1)}^{3}} + C$

$\frac{x^{4}}{6 {(2 x^{4} + 3 x^{2} + 1)}^{3}} + C$

answer is B.

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Detailed Solution

$I = \int \frac{3 x^{13} + 2 x^{11}}{{(2 x^{4} + 3 x^{2} + 1)}^{4}} d x = \int \frac{3 x^{13} + 2 x^{11}}{x^{16} {(2 + \frac{3}{x^{2}} + \frac{1}{x^{4}})}^{4}} d x$
$I = \int \frac{\frac{3}{x^{3}} + \frac{2}{x^{5}}}{{(2 + \frac{3}{x^{2}} + \frac{1}{x^{4}})}^{4}} d x$
Let $2 + \frac{3}{x^{2}} + \frac{1}{x^{4}} = t, - 2 (\frac{3}{x^{3}} + \frac{2}{x^{5}}) d x = d t$
Then, $I = \int \frac{- \frac{d t}{2}}{t^{4}} = - \frac{1}{2} \frac{t^{- 4 + 1}}{- 4 + 1} + C$
$I = \frac{- 1}{2} \times \frac{1}{(- 3)} \frac{1}{{(2 + \frac{3}{x^{2}} + \frac{1}{x^{4}})}^{3}} + C \Rightarrow I = \frac{1}{6} \frac{x^{12}}{{(2 x^{4} + 3 x^{2} + 1)}^{3}} + C$