Integration of Algebraic Functions

Given a constant \(c\) and two functions \(f(x)\) and \(g(x)\), two basic properties of integrals are

\(\begin{array}{rrl} \quad \text{1.} &\displaystyle \int c f(x)\, dx &= c \displaystyle\int f(x)\, dx \\ \quad \text{2.} &\displaystyle \int \left( f(x) + g(x) \right)\, dx & = \displaystyle\int f(x)\, dx + \int g(x)\, dx . \end{array}\)

These two properties follow from the differential formulas:

\[ \begin{align} \frac{d}{dx} cF(x) &= c \frac{d}{dx} F(x) ~\text{ and}\\ \frac{d}{dx} \left( F(x) + G(x) \right) &= \frac{d}{dx} F(x) + \frac{d}{dx} G(x) . \end{align}\]

For a real number \(n\), the indefinite integral of \(f(x)= x^n \) is \[ \int x^n dx = \frac{x ^{n+1}}{n+1} + C,\] where \(C\) is the constant of integration. \(_\square\)

This can easily be shown through an application of the fundamental theorem of calculus.

We know by the power rule that \[\frac{d}{dx} x^m = mx^{m-1},\] where \(m\) is an arbitrary constant. Multiplying both sides by \(\frac{1}{m}\) gives \[\frac{d}{dx} \frac{x^m}{m} = x^{m-1}.\]

Note that \(m≠0\) as this would make the left-hand side indeterminate. The first fundamental theorem of calculus tells us that differentiation is the opposite of integration. Using this fact, let us take the integral of both sides: \[\int x^{m-1}\, dx = \int \frac{d}{dx} \frac{x^m}{m}\, dx = \frac{x^m}{m}+C.\] As stated above, \(m\) is an arbitrary constant, so we can set \(m\) as anything as long as \(m≠0\). Letting \(m=n+1\) where \(n≠-1,\) we have \[\int x^n\, dx = \frac{x^{n+1}}{n+1}+C. \ _\square\]

Evaluate the integral \(\displaystyle \int 2x^3\, dx \).

---

We have \[\int \left(2x^3\right)\, dx = 2\int x^3\, dx. \qquad \text{ (by Property 1)}\] Applying our theorem from above, \[2\int x^3\, dx=2\left(\frac{x^{3+1}}{3+1}\right)+C=\frac{x^4}{2}+C,\] where \(C\) is the constant of integration. \(_\square\)

As stated above, we can let \(n\) be any arbitrary real number as long as \(n≠-1\). This means that our formula works not only for whole numbers, but for negative numbers, rational numbers and irrational numbers as well. For more information on the integral of \(\frac{1}{x}\), please refer to the wiki Integration of Rational Functions.

Evaluate the integral \(\displaystyle \int \frac{7}{x^4}\, dx.\)

---

We have \[\int \frac{7}{x^4}\, dx=\int \left(7x^{-4}\right)\, dx = 7\int x^{-4}\, dx. \qquad \text{ (by Property 1)}\] Applying our theorem from above, \[7\int x^{-4} dx=7\left(\frac{x^{-4+1}}{-4+1}\right)+C=7\left(\frac{x^{-3}}{-3}\right)+C=\frac{-7}{3x^3}+C,\] where \(C\) is the constant of integration. \(_\square\)

Evaluate the integral \(\displaystyle \int \left(\frac{π+1}{11}x^π\right)\, dx \).

---

We have \[\int \left(\frac{π+1}{11}x^π\right)\, dx= \frac{π+1}{11}\int x^π\, dx. \qquad \text{ (by Property 1)}\] Applying our theorem from above, \[\frac{π+1}{11}\int x^π dx=\frac{π+1}{11}\left(\frac{x^{π+1}}{π+1}\right)+C=\frac{x^{π+1}}{11}+C,\] where \(C\) is the constant of integration. \(_\square\)

Our theorem can be extended to all polynomial expressions through the application of the properties of integrals.

By the properties of integrals, given a polynomial \[ f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_ 0, \] we have

\[ \int f(x)\, dx = \frac{a_n}{n+1} x^{n+1} + \frac{ a_{n-1} }{n} x^{n} + \cdots + \frac{a_2}{3} x^3 + \frac{ a_1}{2} x^2 + a_0 x + C,\] where \(C\) is the constant of integration. \(_\square\)

Evaluate the integral \(\displaystyle \int \left( -3x^4 + \pi x^{11} \right)\, dx \).

---

Using the above properties, we have \[ \begin{align} \int \left( -3x^4+\pi x^{11} \right)\, dx &= \int \left(-3x^4\right)\, dx + \int \pi x^{11}\, dx &\qquad \text{ (by Property 2)}\\ &= -3 \int x^4\, dx + \pi \int x^{11}\, dx &\qquad \text{ (by Property 1)}\\ & = \frac{-3}{5} x^5 + \frac{\pi}{12} x^{12} + C, \end{align}\] where \(C\) is the constant of integration. \(_\square\)

Evaluate the integral \(\large \displaystyle \int \frac{ 2 x^{1/3} - 17 x^{-1/3} }{\sqrt{x}} dx.\)

---

We have

\[ \begin{align} \int \frac{ 2 x^{1/3} - 17 x^{-1/3} }{ \sqrt{x} }\, dx &= \int \left( \frac{ 2 x^{1/3} }{ x^{1/2} } - \frac{ 17x^{-1/3} } {x^{1/2} } \right)\, dx\\ &= \int 2 x^{1/3-1/2}\, dx - \int 17 x^{-1/3-1/2}\, dx\\ &= 2 \int x^{-1/6}\, dx- 17 \int x^{-5/6}\, dx\\ &= \frac{2}{ \frac{5}{6} } x^{5/6} - \frac{17}{ \frac{1}{6} } x^{1/6} + C\\ &= \frac{12}{5}x^{5/6} - 102 x^{1/6} + C, \end{align}\] where \(C\) is the constant of integration. \(_\square\)

If \(f(x) = (3x-2) ^4\), what is \(\displaystyle \int f(x)\, dx \)?

---

Note that \(f(x)\) is a polynomial but is not in the form given in the summary above. We will later see methods to integrate this function directly, but to use the basic properties above, we first expand the polynomial by the binomial theorem. This gives

\[\begin{align} f(x) &= (3x-2)^4 \\ &= (3x)^4 + 4(3x)^3(-2) + \frac{4 \cdot 3}{2} (3x)^2 (-2)^2 + \frac{4\cdot 3 \cdot 2}{1 \cdot 2 \cdot 3} (3x)(-2)^3 + (-2)^4\\ &= 81 x^4 - 216 x^3 + 216 x^2 - 96 x + 16, \end{align}\]

which implies

\[\begin{align} \int f(x) dx &= \frac{81}{5} x^5 - \frac{216}{4}x^4 + \frac{216}{3} x^3 - \frac{96}{2} x^2 + 16 x + C\\ & = \frac{81}{5} x^5 - 54x^4 + 72 x^3 - 48 x^2 + 16x + C, \end{align}\] where \(C\) is the constant of integration. \(_\square\)