# Motivation for solving integrals of specific forms.

I was reading a book (Problems in One Variable Calculus by IA Maron) and there was a chapter called Integration of a Binomial Differential .

I'm quoting a part of text

The integral $$\int x^m(a+bx^n)^p dx$$ can be evaluated in the following ways-

Case I. $$p$$ is an integer . Then if $$p>0$$ break the binomial up and if $$p<0$$ then put $$x=t^k$$ where k is the common denominator of the fractions $$m$$ and $$n$$.

Case II. $$\frac{m+1}{n}$$ is an integer. We put $$a+bx^n=t^k$$ where k is the denominator of the fraction p.

Case III. $$\frac{m+1}{n}+p$$ is an integer. We put $$a+bx^n=t^kx^n$$ where k is the denominator of the fraction p.

I am curious about how people came up with these? What were the motivations behind it? What were they thinking when they came up with these substitutions.

• I'm confused. Your integrand is just $x^m$ times a constant. You can pull the constant out of the integral and just integrate $x^m$, no special techniques needed. May 18, 2014 at 4:11
• Sorry for the inconvenience, it has been edited . May 18, 2014 at 4:18

All of the substitutions stem from the observation that we know how to evaluate the integral if the integrand is a rational function - which is a ratio of two polynomials. Everything else is just manipulating the integral to get it into that form. For an excellent demonstration of the methods to integrate any rational function, please see this.

There are $3$ more obvious sorts of substitutions we might try in order to make the integrand a rational function. Substitution 1: $x=t^k$, Substitution 2: $a+bx^n=t^k$, and Substitution 3: $a+bx^n=t^kx^n$. We now examine when each of these substitutions gives us all integer exponents in the integrand, making it a rational function.

Substitution 1: $x=t^k$. This gives $dx=kt^{k-1}dt$, so we have

$$\int x^m(a+bx^n)^pdx=k\int t^{km}(a+bx^{kn})^pt^{k-1}dt$$

Hence we have a rational function if $km+k-1$, $kn$, and $p$ are all integers. This means that we need $p$ to be an integer and we need $k$ to be a multiple of the denominator of $m$ and $k$ to be a multiple of the denominator of $n$. Any such $k$ will do, but the smallest one is the least common denominator of $m$ and $n$, so we generally think of that as the nicest. Note that this gives rise to case 1 as you gave in the question.

Substitution 2: $a+bx^n=t^k$. This gives $bnx^{n-1}dx=kt^{k-1}dt$, and $x=(\frac{1}{b}t^k-\frac{a}{b})^\frac{1}{n}$ so we have $$\int x^m(a+bx^n)^pdx=\frac{k}{bn}\int (\frac{1}{b}t^k-\frac{a}{b})^{\frac{m}{n}-\frac{n-1}{n}}t^{kp}t^{k-1}dt$$

Hence we have a rational function if $k$, $\frac{m}{n}-\frac{n-1}{n}$, and $kp+k-1$ are all integers. In order for this to work, we need $\frac{m}{n}-\frac{n-1}{n}=\frac{m+1}{n}-1$ to be an integer, so $\frac{m+1}{n}$ needs to be an integer. The criteria are met if $k$ is a multiple of the denominator of $p$, and again we use the smallest one - the denominator of $p$ - because it is the simplest. Note that this gives rise to case 2 as you gave in the question.

Substitution 3: $a+bx^n=t^kx^n$. This gives $-nax^{-n-1}dx=kt^{k-1}dt$, and $x=(\frac{1}{a}t^k-\frac{b}{a})^\frac{-1}{n}$ so we have

$$\int x^m(a+bx^n)^pdx=\frac{-k}{an}\int (\frac{1}{a}t^k-\frac{b}{a})^{\frac{-m}{n}-\frac{n+1}{n}-p}t^{kp}t^{k-1}dt$$ Hence we have a rational function if $k$, $\frac{-m}{n}+\frac{n+1}{n}+p$, and $kp+k-1$ are all integers. In order for this to work, we need $\frac{-m}{n}-\frac{n+1}{n}-p=-\frac{m+1}{n}-p-1$ to be an integer, so $\frac{m+1}{n}+p$ needs to be an integer. Completely analogously to the last case, we just need now for $k$ to be a multiple of the denominator of $p$, and the simplest one of those is just the denominator of $p$. Note that this gives rise to case 3 as you gave in the question.

In short the idea is this: Because we know we can deal with rational functions, we search among common substitution types for conditions to turn the integrand into a rational function.

As an aside, a theorem due to Chebyshev says that these cases are the only ones in which the integral can be evaluated in terms of elementary functions.

• That's more than an aside - it leads to the field of integration in finite terms which I find fascinating. Gimme a "C", gimme an "H" gimme an "E-B-Y" - what's that spell? Who's number one? May 24, 2014 at 16:45