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

2 Answers
Apr 10, 2018

#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#

Apr 10, 2018

#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#