How do you integrate #e^-lnx#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Jim H Apr 6, 2015 Use these two facts: #-lnx = ln x^(-1)=ln(1/x)# #e^lnu = u# So, #int x^(-lnx) dx = int 1/x dx=ln absx+C#. Since the original function involves #lnx#, we are justified in assuming that #x>0# and writing: #int x^(-lnx) dx =lnx+C#. Answer link Related questions How do you evaluate the integral #inte^(4x) dx#? How do you evaluate the integral #inte^(-x) dx#? How do you evaluate the integral #int3^(x) dx#? How do you evaluate the integral #int3e^(x)-5e^(2x) dx#? How do you evaluate the integral #int10^(-x) dx#? What is the integral of #e^(x^3)#? What is the integral of #e^(0.5x)#? What is the integral of #e^(2x)#? What is the integral of #e^(7x)#? What is the integral of #2e^(2x)#? See all questions in Integrals of Exponential Functions Impact of this question 32839 views around the world You can reuse this answer Creative Commons License