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

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

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

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

answer is B.

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Detailed Solution

Let

$I = \int_{}^{} \frac{3 x^{13} + 2 x^{11}}{{(2 x^{4} + 3 x^{2} + 1)}^{4}} d x = \int_{}^{} \frac{\frac{3}{x^{3}} + \frac{2}{x^{5}}}{{(2 + \frac{3}{x^{2}} + \frac{1}{x^{4}})}^{- 4}} d x$

[On dividing numerator and denominator by $x^{16}]$

Now, put $2 + \frac{3}{x^{2}} + \frac{1}{x^{4}} = t$

$\Rightarrow (\frac{- 6}{x^{3}} - \frac{4}{x^{5}}) d x = d t \Rightarrow (\frac{3}{x^{3}} + \frac{2}{x^{5}}) d x = - \frac{d t}{2}$

So, $I = \int_{}^{} \frac{- d t}{2 t^{4}} = - \frac{1}{2} \times \frac{t^{- 4 + 1}}{- 4 + 1} + C = \frac{1}{6 t^{3}} + C$

$= \frac{1}{6 {(2 + \frac{3}{x^{2}} + \frac{1}{x^{4}})}^{3}} + C [∵ t = 2 + \frac{3}{x^{2}} + \frac{1}{x^{4}}]$ $= \frac{x^{12}}{6 {(2 x^{4} + 3 x^{2} + 1)}^{3}} + C$