Question

How do you find the integral for `(x)/(sqrt(1+2x))dx`?

Answer 1

I would probably use parts, but since this is asked in Integration by substitution, I'll do it that way.

`int (x)/(sqrt(1+2x))dx`

Let `u = 1+2x`. This makes `du = 2 dx` and `dx= 1/2 du`.

The substitution is not complete until we also note that
`u = 1+2x` makes `x= (u-1)/2`

`int (x)/(sqrt(1+2x))dx = int x 1/(sqrt(1+2x)) *dx`

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

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

You can probably finish this yourself.

`= 1/4[2/3 u^(3/2) - 2/1 u^(1/2)]+C = 1/6[u^(3/2)-3u^(1/2)]+C`

`int (x)/(sqrt(1+2x))dx =1/6[sqrt(1+2x)^3-3sqrt(1+2x)]+C`