What is #f(x) = int 1/x-e^x dx# if #f(1) = 0 #? Calculus Techniques of Integration Evaluating the Constant of Integration 1 Answer Alan N. May 5, 2016 #f(x)=ln(x)-e^x+e# Explanation: #f(x)= int(1/x-e^x)dx# #=int1/xdx-inte^xdx# (Linearity) #int1/xdx = ln(x)# (Standard integral) #inte^x = e^x# (Exponential rule) Therefore: #f(x) = ln(x)-e^x +C# (Where C is the constant of integration) We are told that #f(1)=0# Thus: #ln(1)-e^1+C=0# #0 - e +C =0# #C=e# Hence: #f(x)=ln(x)-e^x+e# 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 1564 views around the world You can reuse this answer Creative Commons License