How to Integrate Sqrt(x^4 + 1) and Sqrt(9x^4 + 1)
Functions of the form f(x) = sqrt(p(x)), where p(x) is a polynomial, are only integrable when the degree of p is 0, 1, or 2, or else if p(x) = q(x)^2 for some other polynomial q(x). This means the functions sqrt(x^4 + 1) and sqrt(9x^4 + 1) are non-integrable functions, meaning their antiderivatives cannot be expressed in terms of elementary functions. The function y = sqrt(9x^4 + 1) arises when computing the arc length of a cubic function, so it is of particular interest to mathematicians.
Applying integration techniques such as trig substitution or integration by parts will help you arrive at related integrals, but these methods will not resolve into an antiderivative function. Nonetheless, it is possible to approximate the antiderivatives of these functions with power series, which will allow you to estimate definite integrals.
Power Series for sqrt(x^4 + 1) and Its Antiderivative
To express the function sqrt(x^4 + 1) as an infinite series, we need to consider it as a power function (x^4 + 1)^(1/2), and then apply the Binomial Theorem. The Binomial Theorem is a formula for expanding products of the form (z + y)^N, for some numbers z and y and some positive integer N. The formula is
(z + y)^N
= ∑(0 ≤ k ≤ N) B(N, k)*(z^k)*(y^(N-k))
where the binomial coefficient B(N, k) is given by B(N, k) = N!/[k!(N-k)!]. We want to use this formula to expand square roots, the case when N = 1/2. The only problem is, how do you define binomial coefficients and factorial functions for non-integer values of N? To do this, we have to rewrite the formula for B(N, k) in an alternative but equivalent form:
Replacing N = 1/2 into the equation above gives us the binomial coeffecients for all values of non-negative integers k:
B(1/2, 0) = 1
B(1/2, 1) = 1/2
B(1/2, 2) = -1/8
B(1/2, 3) = 1/16
B(1/2, 4) = -5/128
The general formula for B(1/2, k), where k is a non-negative integer, is
= [(-1)^(k+1)]*B(2k, k)/[(2k-1)*(4^k)]
We can use this to expand the product (x^4 + 1)^(1/2) as an infinite series letting k run from 0 to infinity. The first few terms of this power series are
(x^4 + 1)^(1/2)
= 1 + (x^4)/2 - (x^8)/8 + (x^12)/16 - 5(x^16)/128 + ... + B(1/2, k)*x^(4k) + ...
Integrating this series term by term gives us
∫ sqrt(x^4 + 1) dx
= x + (x^5)/10 - (x^9)/72 + (x^13)/208 - 5(x^17)/2176 + ...
Thus we have an infinite power series approximation for the antiderivative of sqrt(1 + x^4). Unfortunately, the series expansions of sqrt(x^4 + 1) and its antiderivative only work for values of x that are between -1 and 1 (inclusive). To make this series into an approximate antiderivative, use the first five or six terms. Since it is a convergent power series, the more terms you use, the better the approximation for the specified range of x.
Power Series for sqrt(9x^4 + 1) and Its Antiderivative
Now that we did the hard work of constructing a power series expansion for sqrt(x^4 + 1), we can applying the same method above to find a series for sqrt(9x^4 + 1). All we have to do is replace x^4 with 9x^4. This gives us
sqrt(1 + 9x^4) = (9x^4 + 1)^(1/2)
= 1 + (9x^4)/2 - (81x^8)/8 + (729x^12)/16 - 5(6561x^16)/128 + ...
∫ sqrt(9x^4 + 1) dx
= x + 9(x^5)/10 - 81(x^9)/72 + 729(x^13)/208 - 32805(x^17)/2176 + ...
These series only converge if x is between -1/sqrt(3) and 1/sqrt(3) (inclusive), so they are of limited practical use in most problems. As before, to create an approximate antiderivative function, you simply truncate the series after a finite number of terms. The more terms you inlcude, the better the approximation for the given range of x.
Better Power Series Approximations
Unfortunately the previous two integral series can only be used for a small range of x. But we can construct a more useful power series by writing sqrt(x^4 + 1) in an equivalent form:
sqrt(x^4 + 1)
= x^2 * sqrt(x^4 +1)/x^2
= x^2 * sqrt(1 + (1/x)^4)
Using the work we did above, the power series expansion of sqrt(1 + (1/x)^4) is
sqrt(1 + (1/x)^4)
= 1 + 1/(2x^4) - 1/(8x^8) + 1/(16x^12) - 5/(128x^16) + ...
Multiplying this by x^2, we get another power series for sqrt(x^4 + 1):
sqrt(x^4 + 1)
= x^2 + 1/(2x^2) - 1/(8x^6) + 1/(16x^10) - 5/(128x^14) + ...
Integrating this term by term gives us
∫ sqrt(x^4 + 1) dx
= (x^3)/3 - 1/(2x) + 1/(40x^5) - 1/(144x^9) + 5/(1664x^13) - ...
As luck (or math) would have it, this series converges for all x ≥ 1 and x ≤ -1. And by replacing x^2 with 3x^2 in the steps above, we can construct another power series for sqrt(9x^4 + 1) and its antiderivative as well.
sqrt(9x^4 + 1)
= 3x^2 + 1/(6x^2) - 1/(216x^6) + 1/(3888x^10) - 5/(279936x^14) + ...
∫ sqrt(9x^4 + 1) dx
= x^3 - 1/(6x) + 1/(1080x^5) - 1/(34992x^9) + 5/(3639168x^13) - ...
These two series converge for x ≥ 1/sqrt(3) and x ≤ -1/sqrt(3). Now we have a complete set of series we can use for the full range of x!
Example of Approximate Integration
Find the arc length of the curve y = x^3 from x = 0 to x = 2, shown in the graph below.
To solve this problem, we first apply the arc length integral formula
Arc Length = ∫ sqrt(1 + (y')^2) dx, [0 ≤ x ≤ 2]
= ∫ sqrt(1 + (3x^2)^2) dx, [0 ≤ x ≤ 2]
= ∫ sqrt(1 + 9x^4) dx, [0 ≤ x ≤ 2]
Since we have an infinite series expansion for this integral for values of |x| ≤ 1/sqrt(3) and another for values of |x| ≥ 1/sqrt(3), we need to break this integral up into two pieces. The first piece is
∫ sqrt(1 + 9x^4) dx, [0 ≤ x ≤ 1/sqrt(3)]
and the second piece is
∫ sqrt(1 + 9x^4) dx, [1/sqrt(3) ≤ x ≤ 2]
Their sum gives the entire arc length of the cubic curve along the interval [0, 2]. The first piece requires us to use the first power series expansion for the antiderivative of sqrt(9x^4 + 1). We use the first five terms of the sequence, plug in the end points x = 0 and x = 1/sqrt(3) and subtract. This gives us
∫ sqrt(1 + 9x^4) dx, [0 ≤ x ≤ 1/sqrt(3)] ≈ 0.6285156.
The second integral piece requires us to use the second infinite series for the antiderivative of sqrt(9x^4 + 1). Again, we use the first five terms of the sequence, plug in the endpoints x = 1/sqrt(3) and x = 2, and subract. This gives us
∫ sqrt(1 + 9x^4) dx, [1/sqrt(3) ≤ x ≤ 2] ≈ 7.9967521.
The sum of these two approximations is the approximate arc length of the cubic polynomial
0.628516 + 7.9967521 = 8.6252677.
If you compute the integral with a numerical integrator, you get 8.6303292, so our value is a slight under-estimate.
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