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

3 Answers
Andrea S.
Feb 3, 2017

int (lnx)^2/xdx = (lnx)^3/3+C∫(lnx)2xdx=(lnx)33+C

Explanation:

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

y=lnxy=lnx

dy = (dx)/xdy=dxx

and have:

int (lnx)^2/xdx = int y^2dy = y^3/3+C = (lnx)^3/3+C∫(lnx)2xdx=∫y2dy=y33+C=(lnx)33+C

Gió
Feb 3, 2017

I got: 1/3(ln(x))^3+c13(ln(x))3+c

Explanation:

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

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

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∫(ln(x))2xdx+2∫(ln(x))2xdx=(ln(x))3
add the two integrals and rearrange:
3int(ln(x))^2/xdx=(ln(x))^33∫(ln(x))2xdx=(ln(x))3
int(ln(x))^2/xdx=1/3(ln(x))^3+c∫(ln(x))2xdx=13(ln(x))3+c

mason m
Feb 3, 2017

int(lnx)^2/xdx=1/3(lnx)^3+C∫(lnx)2xdx=13(lnx)3+C

Explanation:

I=int(lnx)^2/xdxI=∫(lnx)2xdx

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

I=int(lnx)^2(1/xdx)=intu^2du=1/3u^3=1/3(lnx)^3+CI=∫(lnx)2(1xdx)=∫u2du=13u3=13(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

Related questions
See all questions in Integration by Parts
Impact of this question
21149 views around the world
You can reuse this answer
Creative Commons License