Question

How do you integrate `int lnx/x` by integration by parts method?

Answer 1

`int lnx /x dx = 1/2(lnx)^2 + c`

Explanation:

The formula for IBP is `int u (dv)/dx dx= uv- int v(du)/dx dx`

Se hopefully we can identify one function which simplifies when differentiated, and the other which simplifies when integrated (or is at least easier to integrate than the original function).

so hopefully it is obvious that we choose:
`u=lnx` and `(du)/dx=1/x`

so `u=lnx=>(du)/dx=1/x`
and `(dv)/dx=1/x=>v=lnx`

`:. int lnx 1/x dx = lnxlnx-int lnx1/xdx + c_1`
`:. int lnx /x dx = (lnx)^2-int lnx/xdx + c_1`
`:. 2int lnx /x dx = (lnx)^2 + c_1`
`:. int lnx /x dx = 1/2(lnx)^2 + 1/2c_1`
`:. int lnx /x dx = 1/2(lnx)^2 + c`

Answer 2

`=>I=(1/2)(lnx)^2+K`

Explanation:

integration by parts formula:

`intu((dv)/dx)dx=uv-intv((du)/dx)dx`

for `color(red)(I=(int(lnx/x)dx))`

Let `u= lnx=>(du)/dx=1/x`

Let `(dv)/dx=1/x=>v=lnx`

`:. I=(lnx)(lnx)-color(red)(int(lnx/x)dx)`

`=>I=(lnx)^2-color(red)(I)+C`

`=>2I=(lnx)^2+C`

`=>I=(1/2)(lnx)^2+K`