Question

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

Answer 1

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#