Solving Hard Antiderivative: u-substitution & Trig Substitution

[pierce15]

1. The problem, the whole problem, and nothing but the problem

\int \frac{dx}{ \sqrt{ 1+ \sqrt{ 1+ \sqrt{ x } } } }

Homework Equations

u-substitution (in the style of trig substitution)
I think that I've got it figured out, I just don't know if my substitutions were legitimate.

The Attempt at a Solution

\int \frac{dx}{ \sqrt{ 1+ \sqrt{ 1+ \sqrt{ x } } } }
u^2 = x, 2u \, du = dx
2 \int \frac{u}{ \sqrt{ 1+ \sqrt{ 1+ u } } } du
v^2 = 1+u, 2v \, dv = du
4 \int \frac{v(v^2 - 1)}{ \sqrt{ 1+ v } } dv

Now just simplifying:

4 \int \frac{v(v+1)(v-1)\sqrt{ 1+ v }}{ v+1 } dv
4 \int v(v-1)\sqrt{ 1+ v } \, dv
w^2 = 1+v, 2w \, dw = dv
8 \int w(w^2 -1 )(w^2 - 2)(w) \, dw
8 \int w^2(w^2 -1 )(w^2 - 2) \, dw

I'm sure that I could foil this out, integrate, and make the 4 or 5 substitutions. Before I do so, is my work justified? If so, is there a trick to the last integral (without foiling it all out)?

Thanks

 

[iRaid]

If you foil it out it becomes w⁶-3w⁴+2w²... You can solve this integral pretty easily I assume?
And it looks good to me

 

[pierce15]

Oh, that's embarrassing, I thought that it was going to be ugly...

= 8 \int w^6 - 3w^4 + 2w^2 dw
= 8 \big[ w^7/7 -3w^5/5 + 2w^3/3 \big]
=8 \big[ (1+v)^{7/2}/7 -3(1+v)^{5/2}/5 + 2(1+v)^{3/2}/3 \big]
=8 \big[ (1+\sqrt{1+u})^{7/2}/7 - 3(1+\sqrt{1+u})^{5/2}/5 + 2(1+\sqrt{1+u})^{3/2}/3 \big]
=8 \big[ \big(1+\sqrt{1+\sqrt{x}}\big)^{7/2}/7 - 3\big(1+\sqrt{1+\sqrt{x}}\big)^{5/2}/5 + 2\big(1+\sqrt{1+\sqrt{x}}\big)^{3/2}/3 \big]

= \frac{ 8\big(1+\sqrt{1+\sqrt{x}}\big)^{7/2}}{7} - \frac{24\big(1+\sqrt{1+\sqrt{x}}\big)^{5/2}}{5} + \frac{16\big(1+\sqrt{1+\sqrt{x}}\big)^{3/2}}{3}

 

[Ray Vickson]

1. The problem, the whole problem, and nothing but the problem

\int \frac{dx}{ \sqrt{ 1+ \sqrt{ 1+ \sqrt{ x } } } }

Homework Equations

u-substitution (in the style of trig substitution)
I think that I've got it figured out, I just don't know if my substitutions were legitimate.

The Attempt at a Solution

\int \frac{dx}{ \sqrt{ 1+ \sqrt{ 1+ \sqrt{ x } } } }
u^2 = x, 2u \, du = dx
2 \int \frac{u}{ \sqrt{ 1+ \sqrt{ 1+ u } } } du
v^2 = 1+u, 2v \, dv = du
4 \int \frac{v(v^2 - 1)}{ \sqrt{ 1+ v } } dv

Now just simplifying:

4 \int \frac{v(v+1)(v-1)\sqrt{ 1+ v }}{ v+1 } dv
4 \int v(v-1)\sqrt{ 1+ v } \, dv
w^2 = 1+v, 2w \, dw = dv
8 \int w(w^2 -1 )(w^2 - 2)(w) \, dw
8 \int w^2(w^2 -1 )(w^2 - 2) \, dw

I'm sure that I could foil this out, integrate, and make the 4 or 5 substitutions. Before I do so, is my work justified? If so, is there a trick to the last integral (without foiling it all out)?

Thanks

I don't know what "foiling out" means, but the final integral is easy.

 

[SammyS]

I don't know what "foiling out" means, but the final integral is easy.

Ray,

FOIL is a mnemonic for multiplying (expanding) two binomials.

The product of the First term of each binomial
 plus
the product of the two Outer terms
 plus
the product of the two Inner terms
 plus
the product of the Last term of each binomial .

\displaystyle (a+b)(c+d) = \underbrace{ac}_\mathrm{first} + \underbrace{ad}_\mathrm{outside} + \underbrace{bc}_\mathrm{inside} + \underbrace{bd}_\mathrm{last}


The terminology has come to be used in a more general (and incorrect) way to refer to expanding the product of two or more polynomials.

 

[SammyS]

1. The problem, the whole problem, and nothing but the problem

\int \frac{dx}{ \sqrt{ 1+ \sqrt{ 1+ \sqrt{ x } } } }

Homework Equations

u-substitution (in the style of trig substitution)
I think that I've got it figured out, I just don't know if my substitutions were legitimate.

The Attempt at a Solution

\int \frac{dx}{ \sqrt{ 1+ \sqrt{ 1+ \sqrt{ x } } } }...
4 \int v(v-1)\sqrt{ 1+ v } \, dv

At this point, the substitution

w = 1+v​

would work well.

It gives \displaystyle \int (w-1)(w-2)(w^{1/2})\, dw\ .

 

[pierce15]

I got the problem a while ago, Samuel.∇