Question

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

Answer 1

`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`

Answer 2

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`

Answer 3

`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`