# What is the antiderivative of the function lnx/square root x?

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### 2 Answers

`int (ln(x))/sqrt(x) dx`

`t=sqrt(x)` so `dt=1/(2sqrt(x))dx=1/(2t)dx` or `2t dt = dx`

`int ln(x)/sqrt(x)=int ln(t^2)/t 2tdt=4 int ln(t)dt`

`u=ln(t)` so `du=dt/t` and `dv=dt` so `v=t`

`int ln(t)dt=t ln(t) - int dt + C = t ln(t) - t + C`

so substituting `t=sqrt(x)` and using `ln(sqrt(x))=1/2 ln(x)`

`int ln(x)/sqrt(x) dx=4(sqrt(x)ln(sqrt(x))-sqrt(x))+C=4sqrt(x)(1/2ln(x)-1)+C=2sqrt(x)(ln(x)-2)+C`

So `int ln(x)/sqrt(t) dx = 2sqrt(x)(ln(x)-2)+C`

To determine the antiderivative of the given function, we'll have to calculate the indefinite integral of the given function:

`int` ln x dx/`sqrt(x)`

We'll integrate by parts using the formula:

`int` udv = uv - `int` vdu

Let u = ln x => du = dx/x

Let dv = dx/`sqrt(x)` => v = 2`sqrt(x)`

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

`int` ln xdx/` ` `sqrt(x)` = ln(x^2)*`sqrt(x)` - 2x^(-1/2 + 1)/(1 - 1/2) + C

`int` ln x dx/`sqrt(x)` = ln(x^2)*`sqrt(x)` - 4 `sqrt(x)` + C

`int` ln x dx/`sqrt(x)` = `sqrt(x)` [ln (x^2) - 4] + C

**The antiderivative of the function is Y = `sqrt(x)` [ln (x^2) - 4] + C**