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

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

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

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

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

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

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

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

$=6-2/3=16/3$

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