Question

How do you find the integral `ln x / x^(1/2)`?

Answer 1

`intlnx/x^(1/2)dx=2sqrtxlnx-4sqrtx+C`

Explanation:

Rewrite the integrand, and integrate by parts:

`intlnx/x^(1/2)dx=intx^(-1/2)lnxdx`

`u=lnx`

`du=x^-1dx`

`dv=x^(-1/2)dx`

`v=intx^(-1/2)dx=2x^(1/2)`

`uv-intvdu=2sqrtxlnx-2intx^(1/2)x^-1dx`

`=2sqrtxlnx-2intx^(-1/2)dx=2sqrtxlnx-4sqrtx+C`

So,

`intlnx/x^(1/2)dx=2sqrtxlnx-4sqrtx+C`

Answer 2

`2sqrtxInx-4sqrtx + C`

Explanation:

Let `u=Inx`
and `dv=x^(-1/2)` SO `v=2sqrtx`

Integration by parts = `uv-int v du`

`int(Inx)/x^(1/2) dx` = `2sqrtxtimes Inx - int 2sqrtx times 1/x`

=`2sqrtxInx-2intx^(-1/2) dx`

=`2sqrtxInx-2times2sqrtx + C`

=`2sqrtxInx-4sqrtx + C`