**NOTE:** `y’` means **the derivative of
**`y`, that is `(dy)/dx`.

`y’=sqrt(2x+1`

So

`y=intsqrt(2x+1)\ dx`

This question requires us to integrate, and in the process, to
find the **constant of integration.**

Put `u = 2x + 1`.

Now `du = 2\ dx`

So

`1/2du=dx`

So

`y=intsqrt(2x+1)\ dx`

`=1/2intu^(1//2)du`

`=1/2xx2/3xxu^(3//2)+K`

`=1/3(2x+1)^(3//2)+K`

Now if `x = 0` then `y = 2` (from the question).

`2=(1^(3//2))/3+K`

and this gives `K = 5/3`.

So

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

is the required function.

Get the Daily Math Tweet!

IntMath on Twitter