How do you integrate #int x/(sqrt(2x-1))dx# from [1,5]?

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Answer 1 · 2016-12-19T14:52:48

The answer is #=16/3#

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

Let's rewrite the function

#x/(sqrt(2x-1))=(1/2(2x-1)+1/2)/(sqrt(2x-1))#

#=1/2*(2x-1)/(sqrt(2x-1))+1/(2sqrt(2x-1))#

#=1/2*sqrt(2x-1)+1/2*1/sqrt(2x-1)#

#int_1^5(xdx)/(sqrt(2x-1))=int_1^5 1/2*sqrt(2x-1)dx+int_1^5 1/2*dx/sqrt(2x-1)#

#= [ 1/2*(2x-1)^(3/2)/(2*3/2)+1/2*(2x-1)^(1/2)/(2*1/2) ]_1^5 #

#= [(2x-1)^(3/2)/6+(2x-1)^(1/2)/2 ]_1^5 #

#=(9/2+3/2)-(1/6+1/2)#

#=6-2/3=16/3#