Question

How do you integrate `(ln x / 3)^3`?

Answer 1

`int(lnx/3)^3dx=(xln^3x)/27-(xln^2x)/9+(2xlnx)/9-(2x)/9+C`

Explanation:

Note that `(lnx/3)^3=ln^3x/27`. From this, we see that

`int(lnx/3)^3dx=1/27intln^3xdx`

Using integration by parts:

`intudv=uv-intvdu`

We let

`u=ln^3x" "=>" "du=(3ln^2x)/xdx`

`dv=(1)dx" "=>" "v=x`

This gives us:

`1/27intln^3xdx=1/27xln^3x-1/27int3ln^2xdx`

`=1/27xln^3x-1/9intln^2xdx`

Integrate `intln^2xdx` similarly (using by parts again):

`u=ln^2x" "=>" "du=(2lnx)/xdx`

`dv=(1)dx" "=>" "v=x`

Thus,

`intln^2x=xln^2x-int2lnxdx`

Combining this and multiplying by `-1/9`, we see that:

`1/27intln^3xdx=(xln^3x)/27-(xln^2x)/9+2/9intlnxdx`

Use integration by parts one last time:

`u=lnx" "=>" "du=1/xdx`

`dv=(1)dx" "=>" "v=x`

Thus,

`intlnxdx=xlnx-intdx=xlnx-x`

Hence,

`1/27intln^3xdx=(xln^3x)/27-(xln^2x)/9+(2xlnx)/9-(2x)/9+C`

Don't forget the constant of integration!