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

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

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

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

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

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Solution

The correct option is B $\frac{x^{12}}{6 (2 x^{4} + 3 x^{2} + 1)^{3}} + C$
$I = \int \frac{3 x^{13} + 2 x^{11}}{(2 x^{4} + 3 x^{2} + 1)^{4}} d x$

$= \int \frac{3 x^{13} + 2 x^{11}}{x^{16} {(2 + \frac{3}{x^{2}} + \frac{1}{x^{4}})}^{4}} d x$

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

Let $2 + \frac{3}{x^{2}} + \frac{1}{x^{4}} = t$
$\Rightarrow - (\frac{6}{x^{3}} + \frac{4}{x^{5}}) d x = d t$
$\Rightarrow (\frac{3}{x^{3}} + \frac{2}{x^{5}}) d x = - \frac{1}{2} d t$

$∴ I = - \frac{1}{2} \int \frac{1}{t^{4}} d t$
$= \frac{1}{6 t^{3}} + C$
$= \frac{x^{12}}{6 (2 x^{4} + 3 x^{2} + 1)^{3}} + C$

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Similar questions

\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}}{(2 x^{4} + 3 x^{2} + 1)^{4}} d x

C

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

C

\int \frac{8 x^{43} + 13. x^{38}}{(x^{13} + x^{5} + 1)^{4}} d x = \frac{1}{3} . \frac{x^{a}}{(x^{b} + x^{c} + 1)^{3}} + C

a, b, c ϵ N, (a > b > c

\int \frac{8 x^{43} + 13. x^{38}}{(x^{13} + x^{5} + 1)^{4}} d x = \frac{1}{3} . \frac{x^{a}}{(x^{b} + x^{c} + 1)^{3}} + C

a, b, c ϵ N, (a > b > c

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