How do you use substitution to integrate `x(sqrt(2x+1))dx.`?

This is better-solved using integration by parts. There is nothing u-substitution-related that is easy to think of quickly, here.

I would choose the function that is easier to differentiate and make go away as `u`, such as `x^n`.

Let:

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

Then, we get:

`int udv = uv - intvdu`

`= x/3(2x+1)^(3//2) - int1/3(2x+1)^(3//2)dx`

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

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

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

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

And now we tack on the constant at the end.

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

`int xsqrt(2x+1) dx`

Let `u` be the radicand (the stuff under the radical).

So `u = 2x+1`, then `x = 1/2(u-1)` and `du = 2 dx`

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

`= 1/2 int 1/2(u-1)u^(1/2)du`

`= 1/4 int (u^(3/2)-u^(1/2)) du`

`= 1/4 (2/5u^(5/2) - 2/3u^(3/2)) +C`

`= 1/10u^(5/2)-1/6u^(3/2)+C`

Simplify as desired before reversing the substitution.

I like

`sqrtu/30 u(3u-5) +C`

So we have

`int xsqrt(2x+1) dx = sqrt(2x+1)/30 (2x+1) (3(2x+1)-5) +C`

`= sqrt(2x+1)/30 (2x+1) (6x-2) +C`

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