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Let $(3-x^2)^3$ be a binomial expression. What is the integral of such expression?

First I tried integration by substitution, because there is a composition of two functions. But$\displaystyle\frac{d}{dx}(3-x^2)=2x$ and I learned that this method only works if the integrand has the derivative of the inner function, multiplied by a constant.

Then I used what I learned about power series:

$$(3-x^2)^3=\binom{3}{0}3^3+\binom{3}{1}3^2(-x^2)+\binom{3}{2}3(-x^2)^2+\binom{3}{3}(-x^2)^3$$

And so,

\begin{align} (3-x^2)^3&=27-27x^2+9x^4-x^6\\ \int(3-x^2)^3\,\mathrm dx&=\int(27-27x^2+9x^4-x^6) dx \end{align}

Finally:

$$\displaystyle\int(3-x^2)^3dx=27x-9x^3+\frac{9}{5}x^5-\frac{1}{7}x^7+C$$

But imagine that the power is $10$, or maybe $20$? There is any way to integrate this kind of expression without expand it? Thanks.

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  • $\begingroup$ I don't think there is a general way. However sometimes shortcuts can be made $\endgroup$
    – Amr
    Jun 15, 2013 at 14:40

2 Answers 2

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Let us generalise a bit. The integral of $(a+bx^n)^m$ is given by (just expanding and integrating term-wise):

$$\int(a+bx^n)^m\,\mathrm dx = \sum_{k=0}^m \binom m ka^{m-k}b^k\frac{x^{kn+1}}{kn+1}$$

Now I have not ever seen a nice form for a general expression like the sum on the right-hand side.

In fact, when I just look at it (for example when handed a similar sum in a different context), I usually think nothing except "Hey, that's the integral of $(a+bx^n)^m$".

For many usages of sums like the one on the right, the integral representation is actually a way of simplifying the problem; if there were nice closed forms currently (widely) known, the mathematical practice would probably en masse use those instead of the integral representation.

This all provides some circumstantial evidence to support the gut feeling that there probably is no general way.

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  • $\begingroup$ I see. The integral of the expression is itself a series. For this particular kind of binomial, this particular kind of series is its integral.Thanks $\endgroup$
    – user24047
    Jun 15, 2013 at 15:00
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There is a standard way to solve similar binomial integrals, called the Chebyshev method. It is based on substitution rules, in which 3 cases are given for the standard binomial expression y= x^m * (a + bx^n)^p where m,n,p <>0 and rational numbers. Case 1) if p is a whole, non zero number and m and n fractions, then use the substiution u=x^s, where s is the lcd of the denominator of m and n Case 2) if (m+1)/n is a whole number, then use the substitution u^s=(a + bx^n), where s is the denominator of p Case 3) if (m+1)/n +p is whole number, then use the substitution u^s=(ax^-n + b), where s is the denominator of p if neither cases apply, the integral is not solvable via substitution.

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