What is #f(x) = int 1/(x+3)-1/x dx# if #f(-2)=-1 #?

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Answer 1 · 2018-07-02T14:47:47

#f(x)=\ln|{x+3}/{x}|+\ln2-1 #

#f(x)=\int (\frac{1}{x+3}-1/x)\ dx#

#f(x)=\int (\frac{1}{x+3})\ dx-\int 1/x\ dx#

#f(x)=\ln|x+3|-\ln|x|+C#

#f(x)=\ln|{x+3}/{x}|+C#

Given that #f(-2)=-1# hence we have

# \ln|{-2+3}/{-2}|+C=-1#

#-\ln2+C=-1#

#C=\ln2-1#

#\therefore f(x)=\ln|{x+3}/{x}|+\ln2-1 #