# Find the indefinite integral int x sqrt((2x-1)) dx using integration by substitution int x sqrt((2x-1)) dx Substitute u = 2x -1  Gives x = (u+1)/2 du = 2x dx  X ( (du)/(dx) = 2 and (dx)/(du) = 1/2 )

int x sqrt((2x-1)) dx

u = 2x - 1  (so (du)/(dx) = 2  not 2x)

giving x = (u+1)/2  and (dx)/(du) = 1/2

Now we have

int x sqrt((2x-1)) dx = int ((u+1)/2) sqrt(u) ((dx)/(du)) du = int (1/2)((u+1)/2) sqrt(u) du

= (1/4) (int u^(3/2) du + int u^(1/2)...

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int x sqrt((2x-1)) dx

u = 2x - 1  (so (du)/(dx) = 2  not 2x)

giving x = (u+1)/2  and (dx)/(du) = 1/2

Now we have

int x sqrt((2x-1)) dx = int ((u+1)/2) sqrt(u) ((dx)/(du)) du = int (1/2)((u+1)/2) sqrt(u) du

= (1/4) (int u^(3/2) du + int u^(1/2) du = (1/4)(2/5u^(5/2) + 2/3u^(3/2)) + "constant"

= 1/20u^(5/2) + 1/6u^(3/2) + "constant"

Substituting u = 2x -1 gives

int x (sqrt(2x-1)) dx = 1/20(2x-1)^(5/2) + 1/6(2x-1)^(3/2) + "constant"` answer

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