What is the antiderivative of (lnx)^2 / x^3(lnx)2x3?

1 Answer

int(lnx)^2/x^3dx=-((2(lnx)^2+2lnx+1))/(4x^2)(lnx)2x3dx=(2(lnx)2+2lnx+1)4x2

Explanation:

Let
y=(lnx)^2 /(x^3)y=(lnx)2x3

intydx=int(lnx)^2 /(x^3)dxydx=(lnx)2x3dx
Let
t=lnxt=lnx

e^t=xet=x

e^2t=x^2e2t=x2

t^2=(lnx)^2t2=(lnx)2

(dt)/dx=1/xdtdx=1x

dt=1/xdxdt=1xdx

intydx=int(lnx)^2 /(x^2)(1/xdx)ydx=(lnx)2x2(1xdx)

=intt^2/e^(2t)dt=t2e2tdt

intydx=intt^2e^(-2t)dtydx=t2e2tdt

integrating by parts

intudv=uv-intvduudv=uvvdu

u=t^2u=t2

du=2tdtdu=2tdt

dv=e^-2tdtdv=e2tdt

v=-1/2e^(-2t)v=12e2t

intt^2e^(-2t)dt=t^2(-1/2e^(-2t))-int(-1/2e^(-2t))(2tdt)t2e2tdt=t2(12e2t)(12e2t)(2tdt)

=-t^2/2e^(-2t)+intte^(-2t)dt=t22e2t+te2tdt

=-t^2/2e^(-2t)+I_1=t22e2t+I1
where

I_1=intte^(-2t)dtI1=te2tdt

integrating by parts

intudv=uv-intvduudv=uvvdu

u=tu=t

du=dtdu=dt

dv=e^(-2t)dtdv=e2tdt

v=-1/2e^(-2t)v=12e2t

intte^(-2t)dt=t(-1/2e^(-2t))-int(-1/2e^(-2t))dtte2tdt=t(12e2t)(12e2t)dt

=-1/2te^(-2t)+1/2(-1/2)e^(-2t)=12te2t+12(12)e2t

intte^(-2t)dt=-1/2te^(-2t)-1/4e^(-2t)te2tdt=12te2t14e2t

I_1=-1/2te^(-2t)-1/4e^(-2t)I1=12te2t14e2t

intt^2e^(-2t)dt=-t^2/2e^(-2t)+I_1t2e2tdt=t22e2t+I1

intt^2e^(-2t)dt=-t^2/2e^(-2t)+(-1/2te^(-2t)-1/4e^(-2t))t2e2tdt=t22e2t+(12te2t14e2t)

intt^2e^(-2t)dt=-t^2/2e^(-2t)-1/2te^(-2t)-1/4e^(-2t)t2e2tdt=t22e2t12te2t14e2t

intt^2e^(-2t)dt=-1/4(2t^2+2t+1)e^-2tt2e2tdt=14(2t2+2t+1)e2t

Replacing

t=lnxt=lnx

e^(-2t)=1/x^2e2t=1x2

int(lnx)^2/x^2(1/x)dx=-1/4(2(lnx)^2+2lnx+1)(1/x^2)(lnx)2x2(1x)dx=14(2(lnx)2+2lnx+1)(1x2)

int(lnx)^2/x^3dx=-1/(4x^2)(2(lnx)^2+2lnx+1)(lnx)2x3dx=14x2(2(lnx)2+2lnx+1)

int(lnx)^2/x^3dx=-((2(lnx)^2+2lnx+1))/(4x^2)(lnx)2x3dx=(2(lnx)2+2lnx+1)4x2