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How do you expand #log ([sqrt(x+1)]/(x^2+1))#?

1 Answer

Mar 22, 2016

#log ((sqrt(x+1))/(x^2 +1))##=1/2 log(x+1)-log(x^2+1)#

Explanation:

#log ((sqrt(x+1))/(x^2 +1))#

#=logsqrt(x+1) - log (x^2+1)#-> use property #log_b(x/y)=log_bx-log_by#

#=log(x+1)^(1/2) -log(x^2+1)#

#=1/2 log(x+1)-log(x^2+1)#

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