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

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

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expand (ln(x+1)-ln(x)ln((x+1)x))/(ln((x+1)x))

expand

\frac{ln ( x + 1 ) - ln ( x ) ln ( ( x + 1 ) x )}{ln ( ( x + 1 ) x )}

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

Solution steps

\frac{ln ( x + 1 ) - ln ( x ) ln ( ( x + 1 ) x )}{ln ( ( x + 1 ) x )}

Expand

(x + 1) x : x^{2} + x

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

Expand

(x + 1) x : x^{2} + x

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

Apply the fraction rule:

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

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

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

Cancel

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

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

\frac{8 b + 3 c}{2 ( 6 + b )}

cos (t - 4) U (t - 4)

((9 x + 2) - \frac{8 x}{7 x + 10}) \cdot (x)

x (x - 3)

(1 - 0.8 cos (x))^{2}

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