Question

How do you integrate `int xsqrt(x-1)` by parts?

Answer 1

`x = cos^2 y`
`sqrt(x-1) = siny`
`I = int cos^2 y siny 2cosy siny (-1)`
`I = int (-2) cos^3ysiny`
`I = 1/2 cos^4 y`

Answer 2

Please see below.

Explanation:

`intxsqrt(x-1) dx`

Let `u = x` and `dv = sqrt(x-1) dx`

so that `du=1dx` and `v = 2/3(x-1)^(3/2)`

`uv-intvdu = 2/3x(x-1)^(3/2) - 2/3 int (x-1)^(3/2) dx`

`= 2/3x(x-1)^(3/2) - 2/3 [2/5 (x-1)^(5/2)] +C`

`= 2/3x(x-1)^(3/2) - 4/15 (x-1)^(5/2) +C`.

Rewrite algebraically to taste. I like the answer above, but others might prefer

`= 2/15[5x(x-1)^(3/2) - 2 (x-1)^(5/2)]+C`

Or

`= 2/15(x-1)^(3/2)(3x+2)+C`

Or some equivalent expression.

Answer 3

Without integration by parts one can see that
`x = x - 1 + 1`
so
`x sqrt(x-1) = (x-1)sqrt(x-1) + sqrt(x-1)`
`int x sqrt(x-1) dx = int (x-1)^(3/2)d(x-1) + int (x-1)^(1/2)d(x-1)`
`int x sqrt(x-1) dx = (x-1)^(5/2)2/5 + (x-1)^(3/2)2/3`