Question

Integrate the following? lnx/x^2 dx

Answer 1

`-(lnx-1)/x+C`

Explanation:

We have:

`intlnx/x^2dx`

Which we can write as:

`intlnx*1/x^2dx`

Integration by parts states that:

`intuvdx`, where `u` and `v` are functions, is equal to:

`uintvdx-intu'(intvdx)dx`

Here, `u=lnx` and `v=1/x^2`

Inputting:

`lnx int1/x^2dx-int(d/dxlnx)(int1/x^2dx)dx`

`lnx(-1/x)-int(1/x)(-1/x)dx`

`-lnx/x-int-1/x^2dx`

`-lnx/x+int1/x^2dx`

`-lnx/x-1/x`

`-(lnx-1)/x`

Add the constant of integration:

`-(lnx-1)/x+C`

Answer 2

`-(lnx+1)/x+C`

Explanation:

Use integration by parts and differentiate `lnx` and integrate `1/x^2`
as following:
the forumla for I.B.P is
`int f(x)g'(x)dx=[f(x)g(x)]-int f'(x)g(x)dx`
and let
`lnx=f(x) and 1/x^2 = g'(x )`
which means
`f'(x)=1/x and g(x)=-1/x`
all that is left now is using the I.B.P formula and solve a trivial integral, and simplify the answer. Hence,
`int lnx/x^2dx=[lnx * (-1/x) ]-int 1/x * (-1/x)dx`
simplifying gives
`=- lnx/x +int 1/x^2dx`
which yields
`=- lnx/x +(-1/x)`
which is the same as
`=- (lnx+1)/x`
dont forget to add the +C
`=- (lnx+1)/x+C`