How do you use substitution to integrate $xsqrt(2x-1)$ ?

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Answer 1 · 2015-10-10T14:50:15

Set $u=2x-1=>x=(u+1)/2$ hence $du=2dx$ so now we have

$int xsqrt(2x-1)dx=int (u+1)/2*sqrtu*(du)/2=1/4 int (usqrtu+sqrtu)du= 1/4*2/15*u^(3/2)*(3u+5)+c=1/15*(2x-1)^(3/2)*(3x+1)+c$

How do you use substitution to integrate xsqrt(2x-1)xsqrt(2x-1)?