Question #09a4c

1 Answer
Justin
Jun 13, 2017

#-1/3sqrt((1-2x)^3)+C#

Explanation:

#int sqrt(1-2x) dx#
Let #u=1-2x#
then #du=-2dx#
i.e. #dx=(du)/-2#
#int sqrt(1-2x) dx#
#=intsqrtu(du)/-2#
#=-1/2intu^(1/2)du#
#=-1/2(u^(3/2)/(3/2))#
#=-1/3sqrt(u^3)#
Substituting #u=1-2x# back into the integral,
#=-1/3sqrt((1-2x)^3)+C#

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