What is expand (1-x(ln(x)+c))/(x^2) ?

The solution to expand (1-x(ln(x)+c))/(x^2) is 1/(x^2)-(ln(x))/x-c/x

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expand (1-x(ln(x)+c))/(x^2)

expand

\frac{1 - x ( ln ( x ) + c )}{x ^{2}}

\frac{1}{x ^{2}} - \frac{ln ( x )}{x} - \frac{c}{x}

Solution steps

\frac{1 - x ( ln ( x ) + c )}{x ^{2}}

Expand

1 - x (ln (x) + c) : 1 - x ln (x) - c x

= \frac{1 - x ln ( x ) - c x}{x ^{2}}

Apply the fraction rule:

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

\frac{1 - x ln ( x ) - c x}{x ^{2}} = \frac{1}{x ^{2}} - \frac{x ln ( x )}{x ^{2}} - \frac{c x}{x ^{2}}

= \frac{1}{x ^{2}} - \frac{x ln ( x )}{x ^{2}} - \frac{c x}{x ^{2}}

Cancel

\frac{x ln ( x )}{x ^{2}} : \frac{ln ( x )}{x}

= \frac{1}{x ^{2}} - \frac{ln ( x )}{x} - \frac{c x}{x ^{2}}

Cancel

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

= \frac{1}{x ^{2}} - \frac{ln ( x )}{x} - \frac{c}{x}

\frac{[ ( x - 1 ) ( x + 3 ) ]}{( x - 2 )}

\frac{( 2 a - s )}{( s - a ) ^{2}}

, 6 (u + 5) u^{4} d u

(2 x + 3) \cdot ln (x) + x + 3

csc ((4 n + 1) \cdot 9 0^{\circ})

What is expand (1-x(ln(x)+c))/(x^2) ?
The solution to expand (1-x(ln(x)+c))/(x^2) is 1/(x^2)-(ln(x))/x-c/x