How do you evaluate #int_1^5x/(2x-1)^(3/2)# using the substitution #x=1/2(u^2+1)#?

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Answer 1 · 2018-07-18T15:03:10

#16/3#

Let #x=1/2(u^2+1)\implies dx=udu#

#\int_1^5 \frac{x}{(2x-1)^{3/2}}dx#

#\int_1^3 \frac{1/2(u^2+1)}{(u^2)^{3/2}}\ udu#

#=1/2\int_1^3\frac{u(u^2+1)}{u^3}\ du#

#=1/2\int_1^3(1+u^{-2})\ du#

#=1/2(u+\frac{u^{-2+1}}{-2+1})_1^3#

#=1/2(u-1/u)_1^3#

#=1/2(3-1/3-1+1/1)#

#=2(8/3)#

#=16/3#