# use integration by parts to integrate integration sign, ln x / square root x, whole equation by dx

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Let x=t^2

then by differenciation;

1= 2t*dt/dx

dx = 2tdt

`int(lnx/sqrtx)dx = int(lnt^2/sqrtt^2)*2tdt`

= `int(2lnt)/t*2tdt`

= `4intlnt dt`

=`4(tlnt-t)+C`

= `4[sqrtx*lnsqrtx-sqrtx]+C`

=` 4sqrtx(lnsqrtx-1)+C`

**Sources:**

int ln(x)/sqrt(x) dx

use LIPET to find to use `u=ln(x)` , `dv = 1/sqrt(x) dx`

This gives `du = 1/x dx` and `v = 2sqrt(x)`

Now apply integration by parts

`int udv = uv - int v du` to get

`int ln(x)/sqrt(x) dx = ln(x) (2sqrt(x)) - int 2sqrt(x) 1/x dx`

`int ln(x)/sqrt(x) dx = 2sqrt(x)ln(x) - 2 int sqrt(x)/x dx`

Since `sqrt(x)/x = x^(1/2)/x^1 = x^(-1/2)`

`int ln(x)/sqrt(x) dx = 2sqrt(x)ln(x) - 2 int x^(-1/2) dx`

And finally we get

`int ln(x)/sqrt(x) dx = 2sqrt(x)ln(x) - 4sqrt(x) + C` which is the same as the answer above but without using a table of integrals.