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