# 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...

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` )

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### 1 Answer

`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**