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How to Integrate Sqrt(1 + Sqrt(x)) and Sqrt(x + Sqrt(x))

Nested radical functions such as

y = sqrt(1 + sqrt(x)) and
y = sqrt(x + sqrt(x))

have graphs that look much like those of simpler radical functions, such as y = x^(1/4) and y = sqrt(x), however, their antiderivatives are much more complicated.

To integrate sqrt(1 + sqrt(x)) you must use two applications of u-substitution. To find the antiderivative of sqrt(x + sqrt(x)) you need to use a combination of algebraic manipulations and u-substitution. Learning how to integrate nested radical functions will allow you to solve more complicated integrals like

∫ sqrt[e^(2x) + e^(2.5x)] dx

∫ sqrt(x^10 + x^8) dx

which can be reduced to nested radical forms via u-substitution.


Graph Comparison: y = x^0.25 and y = sqrt(1 + sqrt(x))

Graphs of y = x^(1/4) in red and y = sqrt(1 + sqrt(x) in blue.
Graphs of y = x^(1/4) in red and y = sqrt(1 + sqrt(x) in blue.

Graph Comarison: y = sqrt(x) and y = sqrt(x + sqrt(x))

Graphs of y = sqrt(x) in purple and y = sqrt(x + sqrt(x)) in green.
Graphs of y = sqrt(x) in purple and y = sqrt(x + sqrt(x)) in green.

Antiderivative of Sqrt(1 + Sqrt(x))

We start with the simpler of the two integrals and make the substitution

x = u^2
dx = 2u du

Now the integral becomes

∫ sqrt(1 + sqrt(x)) dx
= ∫ sqrt(1 + u) 2u du
= 2 * ∫ u*sqrt(1 + u) du

Next we make the change of variables

1 + u = w
u = w - 1
du = dw

This produces

2 * ∫ u*sqrt(1 + u) du
= 2 * ∫ (w - 1)*sqrt(w) dw
= 2 * ∫ w^(3/2) - w^(1/2) dw
= (4/5)w^(5/2) - (4/3)w^(3/2) + c

Making the reverse substitutions w = 1 + u and u = sqrt(x) gives us

(4/5)w^(5/2) - (4/3)w^(3/2) + c
= (4/5)(1 + u)^(5/2) - (4/3)(1 + u)^(3/2) + c
= (4/5)(1 + sqrt(x))^2.5 - (4/3)(1 + sqrt(x))^1.5 + c

The equation in bold is the final form of the antiderivative of y = sqrt(1 + sqrt(x)).

Antiderivative of sqrt(1 + sqrt(x))
Antiderivative of sqrt(1 + sqrt(x))

Antiderivative of Sqrt(x + Sqrt(x))

Finding the integral of sqrt(x + sqrt(x)) begins with the same step as in the previous section. We use the change of variables

x = u^2
dx = 2u du

to produce the new integral

∫ sqrt(x + sqrt(x)) dx
= ∫ sqrt(u^2 + u) 2u du
= 2 * ∫ u*sqrt(u^2 + u) du

At this point, we need to use some algebra to transform the integrand u*sqrt(u^2 + u) into something whose antiderivative is more recognizable. The equivalence we'll use is

Algebra Sidebar

u*sqrt(u^2 + u)

= u*sqrt(u^2 + u + 0.25 - 0.25)

= u*sqrt[(u^2 + u + 0.25) - 0.25]

= u*sqrt[(u + 0.5)^2 - 0.25]

= [u + 0.5 - 0.5]*sqrt[(u+0.5)^2 - 0.25]

= (u+0.5)*sqrt[(u + 0.5)^2 - 0.25)
- 0.5*sqrt[(u + 0.5)^2 - 0.25]

u*sqrt(u^2 + u) =

(u+0.5)*sqrt[(u+0.5)^2 - 0.25]
- 0.5*sqrt[(u+0.5)^2 - 0.5]

which is explained in more detail in the blue box on the right. With this algebraic transformation we obtain the integral equivalence

2 * ∫ u*sqrt(u^2 + u) =

∫ 2(u+0.5)*sqrt[(u+0.5)^2 - 0.25]
- sqrt[(u+0.5)^2 - 0.25] du

It probably seems like we made the situation more complicated, but if we make the change of variables u + 0.5 = w and du = dw, we can create a simpler integral expression:

∫ 2w*sqrt(w^2 - 0.25) - sqrt(w^2 - 0.25) dw

Luckily for us, both pieces of this integral are recognizable. The first half is the derivative of the function (2/3)(w^2 - 0.25)^(3/2) and the second half is an integral we worked out in a previous tutorial. Thus, the full antiderivative is

(2/3)(w^2 - 0.25)^(3/2) - (w/2)sqrt(w^2 - 0.25) + Ln[sqrt(w^2 - 0.25) + w]/8 + c

Now we just need to make the reverse substitution with w = u + 0.5 and u = sqrt(x) to solve for the antiderivative of the original functioin, sqrt(x + sqrt(x)). These reverse substitutions give us

(2/3)(u^2 + u)^1.5 - (u+0.5)*sqrt(u^2 + u)/2 + Ln[sqrt(u^2 + u) + u + 0.5]/8 + c

= (2/3)(x+sqrt(x))^1.5 - (sqrt(x)+0.5)*sqrt(x+sqrt(x))/2
+ Ln[sqrt(x+sqrt(x)) + sqrt(x) + 0.5]/8 + c

As you can see, the above antiderivative formula is much more complicated than that of the previous problem.

Antiderivative of sqrt(x + sqrt(x))
Antiderivative of sqrt(x + sqrt(x))

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Comments 2 comments

PoeticFailosophy profile image

PoeticFailosophy 3 years ago from Cuzco, Peroo

How can I integrate more calculus into my life, aside from writing derivative poetry? Isn't the "power rule" redundant since those with the power always rule? Is it a sin to get tan while you're having sec(x)? Is the antiderivative of sqrt(1 + sqrt(1 + sqrt(x))) equal to (-64/315)[1 + sqrt(1 + sqrt(x))]^(3/2) + (8/9)x[1 + sqrt(1 + sqrt(x))]^(1/2) + (8/63)sqrt(x)[1 + sqrt(1 + sqrt(x))]^(3/2) + c? I'm asking for friend.


calculus-geometry profile image

calculus-geometry 3 years ago from Germany Author

I'm guessing you WolframAlpha'd that...

It's only a sin to have sec(x) with your e^x on a cot. I'll spare you (your friend) the sqrt joke.

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