What is #int(lnx^3)/x^4 dx#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Bill K. Dec 3, 2015 #-(ln(x^3))/(3x^3)-1/(3x^3)+C=-(ln(x))/(x^3)-1/(3x^3)+C# Explanation: Use integration-by-parts with #u=ln(x^3)=3ln(x)# and #dv=1/(x^4)dx=x^(-4)dx# to get #du=3/x# and #v=x^(-3)/(-3)#. Then, since #int\ u\ dv=uv-int\ v\ du#, we get #int\ (ln(x^3))/(x^4)\ dx= int\ (3ln(x))/(x^4)\ dx# #=-(ln(x^3))/(3x^3)-int\ -x^(-4)\ dx=-(ln(x^3))/(3x^3)+int\ x^(-4)\ dx# #=-(ln(x^3))/(3x^3)+(x^(-3))/(-3)+C# #=-(ln(x^3))/(3x^3)-1/(3x^3)+C=-(ln(x))/(x^3)-1/(3x^3)+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 1518 views around the world You can reuse this answer Creative Commons License