What is expand (ln|x|+2*(x-6)*x-1)/x ?

The solution to expand (ln|x|+2*(x-6)*x-1)/x is (ln|x|)/x+2x-12-1/x

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expand (ln|x|+2(x-6)x-1)/x

expand

\frac{ln ∣ x ∣ + 2 \cdot ( x - 6 ) \cdot x - 1}{x}

\frac{ln ∣ x ∣}{x} + 2 x - 12 - \frac{1}{x}

Solution steps

\frac{ln ∣ x ∣ + 2 ( x - 6 ) x - 1}{x}

Expand

ln ∣ x ∣ + 2 (x - 6) x - 1 : ln ∣ x ∣ + 2 x^{2} - 12 x - 1

= \frac{ln ∣ x ∣ + 2 x ^{2} - 12 x - 1}{x}

Apply the fraction rule:

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

\frac{ln ∣ x ∣ + 2 x ^{2} - 12 x - 1}{x} = \frac{ln ∣ x ∣}{x} + \frac{2 x ^{2}}{x} - \frac{12 x}{x} - \frac{1}{x}

= \frac{ln ∣ x ∣}{x} + \frac{2 x ^{2}}{x} - \frac{12 x}{x} - \frac{1}{x}

Cancel

\frac{2 x ^{2}}{x} : 2 x

= \frac{ln ∣ x ∣}{x} + 2 x - \frac{12 x}{x} - \frac{1}{x}

Cancel

\frac{12 x}{x} : 12

= \frac{ln ∣ x ∣}{x} + 2 x - 12 - \frac{1}{x}

g (x) (x = 2 + lo g_{4} (x + 1))

\frac{d}{d s} \frac{1}{( s ( s + 2 ) ^{2} )} \cdot (s + 2)

(2 x - y = 2,) (x + y + 3 = 0)

(23 - 32) (5 + 43)

\frac{s ^{2} + s + 1}{s ^{3} ( s + 1 )}

What is expand (ln|x|+2*(x-6)*x-1)/x ?
The solution to expand (ln|x|+2*(x-6)*x-1)/x is (ln|x|)/x+2x-12-1/x