What is the antiderivative of (lnx)2x3?

1 Answer

(lnx)2x3dx=(2(lnx)2+2lnx+1)4x2

Explanation:

Let
y=(lnx)2x3

ydx=(lnx)2x3dx
Let
t=lnx

et=x

e2t=x2

t2=(lnx)2

dtdx=1x

dt=1xdx

ydx=(lnx)2x2(1xdx)

=t2e2tdt

ydx=t2e2tdt

integrating by parts

udv=uvvdu

u=t2

du=2tdt

dv=e2tdt

v=12e2t

t2e2tdt=t2(12e2t)(12e2t)(2tdt)

=t22e2t+te2tdt

=t22e2t+I1
where

I1=te2tdt

integrating by parts

udv=uvvdu

u=t

du=dt

dv=e2tdt

v=12e2t

te2tdt=t(12e2t)(12e2t)dt

=12te2t+12(12)e2t

te2tdt=12te2t14e2t

I1=12te2t14e2t

t2e2tdt=t22e2t+I1

t2e2tdt=t22e2t+(12te2t14e2t)

t2e2tdt=t22e2t12te2t14e2t

t2e2tdt=14(2t2+2t+1)e2t

Replacing

t=lnx

e2t=1x2

(lnx)2x2(1x)dx=14(2(lnx)2+2lnx+1)(1x2)

(lnx)2x3dx=14x2(2(lnx)2+2lnx+1)

(lnx)2x3dx=(2(lnx)2+2lnx+1)4x2