Question

Question #6f533

Answer 1

`intxsqrt(2x+1)dx = (2x+1)^(3/2)(1/5x-1/15)+C`

Explanation:

Using the integration by parts formula

`intudv = uv-intvdu`

Let
`u = x => du = dx`
`dv = (2x+1)^(1/2)dx => v = 1/3(2x+1)^(3/2)`

`intxsqrt(2x+1)dx = intudv`

`=uv - intvdu`

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

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

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

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

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