Integration using substitution??

Question

#int_1^4 1/(sqrtx(4x-1))dx#
and let #u=sqrtx#
Please help me...

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Answer 1 · 2018-03-27T13:27:37

#I=1/2ln|9/5|#

Here,
#I=int_1^4(color(red)(1)/color(red)(sqrtxcolor(blue)((4x-1))))color(red)(dx#

Let #sqrtx=u=>1/(2sqrtx)dx=du=>color(red)(1/sqrtxdx=2du#

#and x=u^2=>x=1tou=1,x=4tou=2#

#I=int_1^2(2du)/(4u^2-1)#

#I=int_1^2(2)/((2u-1)(2u+1))du#

#I=int_1^2((2u+1)-(2u-1))/((2u+1)(2u-1))du#

#I=int_1^2[1/(2u-1)-1/(2u+1)]du#

#=int_1^2 1/2[2/(2u-1)-2/(2u+1)]du#

#=1/2[ln|2u-1|-ln|2u+1|]_1^2#

#=1/2[(ln3-ln5)-(ln1-ln3)]#

#=1/2[ln3-ln5-0+ln3]#

#=1/2ln((3xx3)/5)#

#=1/2ln(9/5)#