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

1 Answer

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

Explanation:

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