What is f(x) = int 1/(x+3)-1/x dx if f(-2)=-1 ? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Harish Chandra Rajpoot Jul 2, 2018 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 Answer link Related questions How do you find the constant of integration for intf'(x)dx if f(2)=1? What is a line integral? What is f(x) = int x^3-x if f(2)=4 ? What is f(x) = int x^2+x-3 if f(2)=3 ? What is f(x) = int xe^x if f(2)=3 ? What is f(x) = int x - 3 if f(2)=3 ? What is f(x) = int x^2 - 3x if f(2)=1 ? What is f(x) = int 1/x if f(2)=1 ? What is f(x) = int 1/(x+3) if f(2)=1 ? What is f(x) = int 1/(x^2+3) if f(2)=1 ? See all questions in Evaluating the Constant of Integration Impact of this question 1318 views around the world You can reuse this answer Creative Commons License