How do you integrate # ln x / x^(1/2)#?

Question

Charles Arocha · Accepted Answer

#= 2 sqrt x ln x - 4 sqrt x + C#

Use IBP

#int u v' = uv - int u'v#

here

#u = ln x, u' = 1/x# #v' = x^{-1/2}, v = 2 x^{1/2}#, using the power rule

so we have

#2 sqrt x ln x - 2 int dx qquad 1/x sqrt x#

#= 2 sqrt x ln x - 2 int dx qquad x^{-1/2}#

#= 2 sqrt x ln x - 2*x^{1/2}/(1/2) + C#

#= 2 sqrt x ln x - 4 sqrt x + C#

Emma Aldridge · Answer

undefined To integrate $ \frac{\ln x}{\sqrt{x}} $, you can use integration by parts. Let $ u = \ln x $ and $ dv = x^{-1/2} \, dx $.

$$ du = \frac{1}{x} \, dx $$
$$ v = \frac{x^{1/2}}{1/2} = 2x^{1/2} $$

Now, apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substituting the values:

$$ \int \frac{\ln x}{\sqrt{x}} \, dx = 2x^{1/2} \ln x - \int 2x^{1/2} \frac{1}{x} \, dx $$

$$ = 2x^{1/2} \ln x - 2 \int x^{1/2 - 1} \, dx $$

$$ = 2x^{1/2} \ln x - 2 \int x^{-1/2} \, dx $$

$$ = 2x^{1/2} \ln x - 2(2x^{1/2}) + C $$

$$ = 2x^{1/2} (\ln x - 1) + C $$

Where $ C $ is the constant of integration.