How do you find the integral #ln(x)/sqrtx#?

Question

5805 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-28T20:04:44

#int (ln(x))/sqrt(x)\ dx=2sqrt(x)ln(x)-4sqrt(x)+C#

Use integration-by-parts. Let #u=ln(x)# so that #du=1/x\ dx# and #dv=1/sqrt(x)\ dx=x^{-1/2}\ dx# so that #v=2x^{1/2}=2sqrt(x)#.

The integration-by-parts formula can be written in abstract form as: #int u\ dv=uv-int v\ du#.

For this problem, this becomes:

#int (ln(x))/sqrt(x)\ dx=2sqrt(x)ln(x)-int 2/sqrt(x)\ dx#

#=2sqrt(x)ln(x)-int 2x^{-1/2}\ dx#

#=2sqrt(x)ln(x)-4x^{1/2}+C#

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