Prove #int_1^3x/(2x-1)^3dx# = 8/25 , Using substitution u = 2x - 1 ?

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Answer 1 · 2018-03-08T15:36:19

See the explanation below

As suggested, let

#u=2x-1#, #=>#, #du=2dx#

#2x=u+1#

#x=(u+1)/2#

Start by calculating the indefinite integral

Therefore,

#int(xdx)/(2x-1)^3=1/4int((u+1)du)/u^3#

#=1/4int(1/u^2+1/u^3)du#

#=1/4(-1/u-1/(2u^2))#

#=1/4(-1/(2x-1)-1/(2(2x-1)^2))+C#

Now, calculate the definite integral

#int_1^3(xdx)/(2x-1)^3=[-1/4*1/(2x-1)-1/8*1/(2x-1)^2)]_1^3#

#=(-1/20-1/200)-(-1/4-1/8)#

#=3/8-11/200#

#=64/200#

#=8/25#