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=∫t2e−2tdt
integrating by parts
∫udv=uv−∫vdu
u=t2
du=2tdt
dv=e−2tdt
v=−12e−2t
∫t2e−2tdt=t2(−12e−2t)−∫(−12e−2t)(2tdt)
=−t22e−2t+∫te−2tdt
=−t22e−2t+I1
where
I1=∫te−2tdt
integrating by parts
∫udv=uv−∫vdu
u=t
du=dt
dv=e−2tdt
v=−12e−2t
∫te−2tdt=t(−12e−2t)−∫(−12e−2t)dt
=−12te−2t+12(−12)e−2t
∫te−2tdt=−12te−2t−14e−2t
I1=−12te−2t−14e−2t
∫t2e−2tdt=−t22e−2t+I1
∫t2e−2tdt=−t22e−2t+(−12te−2t−14e−2t)
∫t2e−2tdt=−t22e−2t−12te−2t−14e−2t
∫t2e−2tdt=−14(2t2+2t+1)e−2t
Replacing
t=lnx
e−2t=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