Question 1094746
1/5 ln(x + 2)^5 + 1/2 [ln(x) − ln(x^2 + 3x + 2)^2]
The 1/5 just cancels the exponent 5.
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= ln(x + 2) + 1/2 [ln(x) − ln(x^2 + 3x + 2)^2] 
= ln(x + 2) + 1/2ln(x) − ln(x^2 + 3x + 2) 
= ln(x + 2) + 1/2ln(x) − ln[(x + 2)*(x + 1)]
= ln(x + 2) + 1/2ln(x) − ln(x + 2) - ln(x + 1)
= 1/2ln(x) - ln(x + 1)
= ln(sqrt(x)) - ln(x + 1)
= {{{ln(sqrt(x)/(x+1))}}}