What is the antiderivative of #(ln(x))^3#?

1 Answer
Mar 14, 2017

#x((ln(x))^3-3(ln(x))^2+6ln(x)-6)+C#

Explanation:

#I=int(ln(x))^3dx#

Let #t=ln(x)#. This implies that #x=e^t# so #dx=e^tdt#. Then:

#I=intt^3(e^tdt)#

On this, we can perform integration by parts a number of times.

#{(u=t^3,=>,du=3t^2dt),(dv=e^tdt,=>,v=e^t):}#

#I=t^3e^t-int3t^2(e^tdt)#

#{(u=3t^2,=>,du=6tdt),(dv=e^tdt,=>,v=e^t):}#

#I=t^3e^t-(3t^2e^t-int6t(e^tdt))#

#I=t^3e^t-3t^2e^t+int6t(e^tdt)#

#{(u=6,=>,du=6dt),(dv=e^tdt,=>,v=e^t):}#

#I=t^3e^t-3t^2e^t+6te^t-int6e^tdt#

#I=t^3e^t-3t^2e^t+6te^t-6e^t+C#

#I=e^t(t^3-3t^2+6t-6)+C#

Using #t=ln(x)# and #e^t=x#:

#I=x((ln(x))^3-3(ln(x))^2+6ln(x)-6)+C#