What is #int(lnx^3)/x^4 dx#?

1 Answer
Dec 3, 2015

Answer:

#-(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#