How do you integrate #int(lnx)^2/x# by integration by parts method?

3 Answers
Feb 3, 2017

#int (lnx)^2/xdx = (lnx)^3/3+C#

Explanation:

Integration by parts is not the best way to solve this integral.
As #d(lnx) = dx/x# we can substitute:

#y=lnx#

#dy = (dx)/x#

and have:

#int (lnx)^2/xdx = int y^2dy = y^3/3+C = (lnx)^3/3+C#

Feb 3, 2017

I got: #1/3(ln(x))^3+c#

Explanation:

We have:
#int(ln(x))^2/xdx=int1/x(ln(x))^2dx=# by parts:
#ln(x)(ln(x))^2-intln(x)*2ln(x)*1/xdx=#
#=(ln(x))^3-2int(ln(x))^2/xdx#
so that basically we have that:

#int(ln(x))^2/xdx=(ln(x))^3-2int(ln(x))^2/xdx#

now a trick...
take the last integral to the left of the equal sign as in a normal equation:
#int(ln(x))^2/xdx+2int(ln(x))^2/xdx=(ln(x))^3#
add the two integrals and rearrange:
#3int(ln(x))^2/xdx=(ln(x))^3#
#int(ln(x))^2/xdx=1/3(ln(x))^3+c#

Feb 3, 2017

#int(lnx)^2/xdx=1/3(lnx)^3+C#

Explanation:

#I=int(lnx)^2/xdx#

Integration by parts is not necessary. The quickest way to do this is with the substitution #u=lnx# which implies that #du=1/xdx#. Then:

#I=int(lnx)^2(1/xdx)=intu^2du=1/3u^3=1/3(lnx)^3+C#

We can do integration by parts, however, letting:

#{(u=(lnx)^2,=>,du=(2lnx)/xdx),(dv=1/xdx,=>,v=lnx):}#

Then:

#I=uv-intvdu#

#I=(lnx)^3-2int(lnx)^2/xdx#

This is the original integral:

#I=(lnx)^3-2I#

#3I=(lnx)^3#

#I=1/3(lnx)^3+C#