Question

What is the antiderivative of `(sqrt(1 + sqrt((1+sqrt(x)))dx`?

Answer 1

`8/9(sqrt(sqrtx+1)+1)^(9/2)-24/7(sqrt(sqrtx+1)+1)^(7/2)+16/5(sqrt(sqrtx+1)+1)^(5/2)+C`

Explanation:

Rearrange the integrand:

`intsqrt(1+sqrt(1+sqrtx))dx=intsqrt(sqrt(sqrtx+1)+1)dx`

Substitute and let `u=sqrt(sqrtx+1)+1`. Finding `du` is a little odd, but it's `du=1/(4sqrt(sqrtx+1)sqrtx)`.

Rearranging our integrand so that our `du` value is present:

`=4int(sqrt(sqrt(sqrtx+1)+1)*sqrt(sqrtx+1)sqrtx)/(4sqrt(sqrtx+1)sqrtx)dx`

Here, make the following substitutions in the numerator:

• `sqrt(sqrt(sqrtx+1)+1)=u^(1/2)`
• `sqrt(sqrtx+1)=u-1`
• `sqrtx=(u-1)^2-1=u(u-2)`

The `1/(4sqrt(sqrtx+1)sqrtx)dx` will become `du`.

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

Expand and integrate!

`=4int(u^2-3u+2)u^(3/2)du=4int(u^(7/2)-3u^(5/2)+2u^(3/2))du`

`=4(2/9u^(9/2)-6/7u^(7/2)+4/5u^(5/2))+C`

`=8/9(sqrt(sqrtx+1)+1)^(9/2)-24/7(sqrt(sqrtx+1)+1)^(7/2)+16/5(sqrt(sqrtx+1)+1)^(5/2)+C`

Answer 2

After... what, 30 minutes? I got `8/9(sqrt(1+sqrt(1 + sqrtx)))^9 - 24/7(sqrt(1+sqrt(1 + sqrtx)))^7 + 16/5(sqrt(1+sqrt(1 + sqrtx)))^5 + C`. See, it's not impossible. Just keep track of your substitutions.

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For

`int sqrt(1+sqrt(1+sqrtx))dx,`

let's try letting `u = sqrtx`. Then, `du = 1/(2sqrtx)dx`, and `2udu = dx`.

`= 2int usqrt(1+sqrt(1+u))du`

Next, try `s = u+1`. Thus, `ds = du`, and we have

`= 2int (s-1)sqrt(1+sqrts)ds`

`= 2int ssqrt(1+sqrts)ds - 2int sqrt(1+sqrts)ds`

Then, we can try letting `t = sqrts`. So, `dt = 1/(2sqrts)ds`, and `2tdt = ds`. So:

`sqrt(1 + sqrts) = sqrt(1+t)`
`s = t^2,` and

`=> 4int t^3sqrt(1+t)dt - 4int tsqrt(1+t)dt`

Now, for one last substitution... Let `r = sqrt(1+t)`. Then, `dr = 1/(2sqrt(1+t))dt` and `2rdr = dt`. Common theme, here.

So now:

`r^2 = 1 + t => t^3 = (r^2 - 1)^3`
`t = r^2 - 1`

`=> 8int (r^2 - 1)^3r^2dr - 8int (r^2 - 1)r^2dr`

This is definitely doable. Expand the polynomial to get:

`=> 8int (r^4 - r^2)(r^2 - 1)(r^2 - 1)dr - 8int r^4 - r^2dr`

`= 8int (r^4 - r^2)(r^4 - 2r^2 + 1)dr - 8int r^4 - r^2dr`

`= 8int r^8 - 2r^6 + r^4 - r^6 + 2r^4 - r^2dr - 8int r^4 - r^2dr`

`= 8int r^8 - 3r^6 + 3r^4 - r^2dr - 8int r^4 - r^2dr`

Hence, we should get:

`= 8[r^9/9 - 3/7r^7 + 3/5r^5 - r^3/3] - 8[r^5/5 - r^3/3]`

`= 8/9r^9 - 24/7r^7 + 24/5r^5 cancel(- 8/3r^3) - 8/5r^5 + cancel(8/3r^3)`

`= 8/9r^9 - 24/7r^7 + 16/5r^5`

Finally, let's back-substitute to get our answer.

`r = sqrt(1+t)`, so our next step gives us

`= 8/9(sqrt(1+t))^9 - 24/7(sqrt(1+t))^7 + 16/5(sqrt(1+t))^5.`

Now, since `t = sqrts`, we have

`= 8/9(sqrt(1+sqrts))^9 - 24/7(sqrt(1+sqrts))^7 + 16/5(sqrt(1+sqrts))^5.`

Then, since `s = u + 1`, we have

`= 8/9(sqrt(1+sqrt(u+1)))^9 - 24/7(sqrt(1+sqrt(u+1)))^7 + 16/5(sqrt(1+sqrt(u+1)))^5.`

Lastly, we first let `u = sqrtx`. So, we end up with:

`= color(blue)(8/9(sqrt(1+sqrt(1 + sqrtx)))^9 - 24/7(sqrt(1+sqrt(1 + sqrtx)))^7 + 16/5(sqrt(1+sqrt(1 + sqrtx)))^5 + C)`

Oh, look at that. It worked.