Question

integrate (3x^13+2x^11)/((2x^4+3x^2+1)^4)​

Answer 1

Step-by-step explanation:

We have,

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

$= \int \frac{3 {x}^{13} + 2 {x}^{11} }{( {x}^{4} (2 + 3 {x}^{ - 2} + {x}^{ - 4} ))^{4} } dx$

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

$= \int \frac{ {x}^{16}( 3 {x}^{ - 3} + 2 {x}^{ - 5}) }{ {x}^{16} (2 + 3 {x}^{ - 2} + {x}^{ - 4} )^{4} } dx$

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

$Let \: 2 + 3 {x}^{ - 2} + {x}^{ - 4} = y \\ \implies - 6 {x}^{ - 3} - 4 {x}^{ - 5} dx = dy \\ \implies - 2(3 {x}^{ - 3} + 2 {x}^{ - 5} )dx = dy$

Now,

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

$= - \frac{1}{2} \int \frac {dy }{ (y )^{4} }$

$= - \frac{1}{2} \int {y}^{ - 4} dy$

$= - \frac{1}{2} \frac{ {y}^{ - 3}}{ - 3} + C$

$= \frac{1}{6} {y}^{ - 3} + C$

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

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