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

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

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

expand

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

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

Solution steps

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

Remove parentheses:

(a) = a

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

Expand

(x + 2) ln (x + 2) : x ln (x + 2) + 2 ln (x + 2)

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

csc^{2} (x) (1 - sin (2 x))

a (a - b)

- 2 (x ln (x) - x)

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

(10 - 5 e^{- x})^{2}

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