What is #int_0^(1/9)1/(1-2sqrt(x))#?

Can someone help me to solve this equation. I came to this:
#int_0^(1/3)1/(1-2u) * 2sqrt(x) du #
with u = #sqrt(x)#
This is:
#2int_0^(1/3)u/(1-2u)du#
and then i don't know what to do.

2 Answers
Cem Sentin
Oct 31, 2017

#1/2*ln(3)-1/3#

Explanation:

#int_0^(1/3) (2u*du)/(1-2u)#

#=-int_0^(1/3) (-2u*du)/(1-2u)#

#=-int_0^(1/3) ((1-2u-1)*du)/(1-2u)#

#=-int_0^(1/3) ((1-2u)*du)/(1-2u)#+#int_0^(1/3) (du)/(1-2u)#

#=-int_0^(1/3) du#+#int_0^(1/3) (du)/(1-2u)#

#=-[u]_0^(1/3)#-#1/2*[ln(1-2u)]_0^(1/3)#

=#-(1/3-0)#-#1/2*[ln(1/3)-Ln(1)]#

=#-1/3-1/2*ln(1/3)#

=#-1/3-1/2*(-ln(3))#

=#1/2*ln(3)-1/3#

Douglas K.
Oct 31, 2017

Starting with your end point:

#2int_0^(1/3) u/(1-2u)du#

Move the 2 back inside the integral as #(-2)/-1#:

#int_0^(1/3) (-2u)/(2u-1)du#

Add zero to the numerator in the form #+1-1#

#int_0^(1/3) (-2u+1 - 1)/(2u-1)du#

Separate into two fractions:

#int_0^(1/3) (-2u+1)/(2u-1) - 1/(2u-1)du#

The first fraction becomes -1:

#int_0^(1/3) -1 - 1/(2u-1)du = [ -u +1/2ln|2u-1|]_0^(1/3)#

#int_0^(1/3) -1 - 1/(2u-1)du = -1/3+ln(3)/2 ~~ 0.22#

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