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

Question

Emilia Aguilar · Accepted Answer

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

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#

Ezra Adams · Answer

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

Isaiah Allen · Answer

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

#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#

Luke Alvarez · Answer

undefined To integrate ∫(ln(x))^2 / x dx using the integration by parts method, you would choose u = ln(x)^2 and dv = dx/x. Then, you would differentiate u to find du and integrate dv to find v.

Following this, you would apply the integration by parts formula:

∫u dv = uv - ∫v du

Substitute the values of u, v, du, and dv into this formula and evaluate to find the integral.