What is #int ln x / sqrtx dx#?

1 Answer
Jim S
May 3, 2018

#2sqrtx(lnx-2)+c#

Explanation:

#intlnx/sqrtxdx=intlnx/color(blue)(2sqrtx)(2color(blue)(dx))#

Substitute #sqrtx=u#

  • #x=u^2#

  • #1/(2sqrtx)dx=du#

#=# #intlnu^color(red)(2)/cancel(u)(2cancel(u)du)# #=#

#int2*2lnudu# #=#

#4intlnudu# #=#

#4(u##lnu##-u)+c# #=#

#4(sqrtxlnsqrtx-sqrtx)+c# #=#

#4(1/2sqrtxlnx-sqrtx)+c# #=#

#2sqrtx(lnx-2)+c# ,

#c##in##RR#

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