What is expand 1/((k+1)ln(k+1)) ?

The solution to expand 1/((k+1)ln(k+1)) is 1/(kln(k+1)+ln(k+1))

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expand 1/((k+1)ln(k+1))

expand

\frac{1}{( k + 1 ) ln ( k + 1 )}

\frac{1}{k ln ( k + 1 ) + ln ( k + 1 )}

Solution steps

\frac{1}{( k + 1 ) ln ( k + 1 )}

Expand

(k + 1) ln (k + 1) : k ln (k + 1) + ln (k + 1)

= \frac{1}{k ln ( k + 1 ) + ln ( k + 1 )}

\frac{x y}{y ^{- 1}}

\frac{( 4 x ^{3} y ^{3} - 16 x ^{4} - 2 x y )}{- 2 x y}

\frac{2 ( - 27 t ^{\frac{13}{12}} + 13 t ^{\frac{9}{4}} )}{351}

\frac{1}{2} (x - 10) (30 - x)

What is expand 1/((k+1)ln(k+1)) ?
The solution to expand 1/((k+1)ln(k+1)) is 1/(kln(k+1)+ln(k+1))