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 n constant of integration)

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

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

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

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

answer is C.

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Detailed Solution

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

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

$\begin{aligned} ∴ (\frac{- 6}{x^{3}} - \frac{4}{x^{5}}) d x = d z \Rightarrow (\frac{3}{x^{3}} + \frac{2}{x^{5}}) d x = \frac{- d z}{2} \\ ∴ I = \frac{- 1}{2} \int \frac{d z}{z^{4}} = \frac{- 1}{2} (\frac{z^{- 3}}{- 3}) + C = \frac{1}{6 z^{3}} + C \\ = \frac{1}{6 {(2 + \frac{3}{x^{2}} + \frac{1}{x^{4}})}^{3}} + C = \frac{x^{12}}{6 {(2 x^{4} + 3 x^{2} + 1)}^{3}} + C \end{aligned}$