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

The solution to expand-((t^2+1)^2)/(4t) is -(t^3)/4-t/2-1/(4t)

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expand-((t^2+1)^2)/(4t)

expand

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

- \frac{t ^{3}}{4} - \frac{t}{2} - \frac{1}{4 t}

Solution steps

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

(t^{2} + 1)^{2} = t^{4} + 2 t^{2} + 1

= - \frac{t ^{4} + 2 t ^{2} + 1}{4 t}

Apply the fraction rule:

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

\frac{t ^{4} + 2 t ^{2} + 1}{4 t} = - (\frac{t ^{4}}{4 t}) - (\frac{2 t ^{2}}{4 t}) - (\frac{1}{4 t})

= - (\frac{t ^{4}}{4 t}) - (\frac{2 t ^{2}}{4 t}) - (\frac{1}{4 t})

Remove parentheses:

(a) = a

= - \frac{t ^{4}}{4 t} - \frac{2 t ^{2}}{4 t} - \frac{1}{4 t}

Cancel

\frac{t ^{4}}{4 t} : \frac{t ^{3}}{4}

= - \frac{t ^{3}}{4} - \frac{2 t ^{2}}{4 t} - \frac{1}{4 t}

Cancel

\frac{2 t ^{2}}{4 t} : \frac{t}{2}

= - \frac{t ^{3}}{4} - \frac{t}{2} - \frac{1}{4 t}

p oss ib l e \frac{1}{3} (3 j + 6)

\frac{( ( 4 m - 4 ) ^{2} )}{m - 1}

5 cos (2 (x + \frac{π}{3}))

(2 x + y) (cos (t)) + (2 y + x) (e^{t})

\frac{5 ( x - 3 ) ^{5}}{16}

What is expand-((t^2+1)^2)/(4t) ?
The solution to expand-((t^2+1)^2)/(4t) is -(t^3)/4-t/2-1/(4t)