What is expand ((2y^4+2y^2+1)(2y^4-2y^2+1)dy)/(4y^4) ?

The solution to expand ((2y^4+2y^2+1)(2y^4-2y^2+1)dy)/(4y^4) is y^4dy+(dy)/(4y^4)

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expand ((2y^4+2y^2+1)(2y^4-2y^2+1)dy)/(4y^4)

expand

\frac{( 2 y ^{4} + 2 y ^{2} + 1 ) ( 2 y ^{4} - 2 y ^{2} + 1 ) d y}{4 y ^{4}}

y^{4} d y + \frac{d y}{4 y ^{4}}

Solution steps

\frac{( 2 y ^{4} + 2 y ^{2} + 1 ) ( 2 y ^{4} - 2 y ^{2} + 1 ) d y}{4 y ^{4}}

Expand

(2 y^{4} + 2 y^{2} + 1) (2 y^{4} - 2 y^{2} + 1) d y : 4 y^{8} d y + d y

= \frac{4 y ^{8} d y + d y}{4 y ^{4}}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{4 y ^{8} d y + d y}{4 y ^{4}} = \frac{4 y ^{8} d y}{4 y ^{4}} + \frac{d y}{4 y ^{4}}

= \frac{4 y ^{8} d y}{4 y ^{4}} + \frac{d y}{4 y ^{4}}

Cancel

\frac{4 y ^{8} d y}{4 y ^{4}} : y^{4} d y

= y^{4} d y + \frac{d y}{4 y ^{4}}

(x - i)^{4}

e^{x} (e^{x} + 1)

[(x^{3} - 2 x) \cdot tan (x)]

\frac{21}{x ( x ^{2} + 7 )}

A (- 31) y B (- 2, - 4)

What is expand ((2y^4+2y^2+1)(2y^4-2y^2+1)dy)/(4y^4) ?
The solution to expand ((2y^4+2y^2+1)(2y^4-2y^2+1)dy)/(4y^4) is y^4dy+(dy)/(4y^4)