Indefinite Integral of Algebraic Functions
Smaridasa Mathematics – Previously we discussed the definition of the integral as the antiderivative or inverse of the derivative of a function. The indefinite integral was also mentioned in the discussion "What is the difference between definite and indefinite integrals". In this article, we will discuss the topic of Indefinite Integral of Algebraic Functions in more depth. From the definition of the integral, if the function \( f(x) \) is the derivative of \( F(x) + c \), then we can write the integral form: \[ \int f(x) \, dx = F(x) + c \]. This article will also cover the properties of indefinite integrals.
Formula for Integrating Algebraic Functions
For a rational number \( n \) with \( n \neq -1 \), and \( a, c \) are real numbers, the following rules apply:
| Function | Indefinite Integral |
|---|---|
| \( \displaystyle \int x^n \, dx \) | \( \displaystyle \frac{1}{n+1} x^{n+1} + c \) |
| \( \displaystyle \int a x^n \, dx \) | \( \displaystyle \frac{a}{n+1} x^{n+1} + c \) |
| \( \displaystyle \int x^{-1} \, dx = \int \frac{1}{x} \, dx \) | \( \displaystyle \ln |x| + c \) |
| \( \displaystyle \int a x^{-1} \, dx = \int \frac{a}{x} \, dx \) | \( \displaystyle a \ln |x| + c \) |
The function \( \ln x \) is read "natural log of x" and is the logarithm with base \( e \approx 2.71828... \).
Example 1: Basic Algebraic Integrals
Determine the integrals of the following:
a) \( \displaystyle \int x^3 \, dx \)
Solution for (a):
Here \( n = 3 \). Using the formula \( \int x^n \, dx = \frac{1}{n+1} x^{n+1} + c \):
\[ \int x^3 \, dx = \frac{1}{3+1} x^{3+1} + c = \frac{1}{4} x^4 + c \]
b) \( \displaystyle \int 3x^2 \, dx \)
Solution for (b):
Here \( a = 3, n = 2 \). Using the formula \( \int a x^n \, dx = \frac{a}{n+1} x^{n+1} + c \):
\[ \int 3x^2 \, dx = \frac{3}{2+1} x^{2+1} + c = \frac{3}{3} x^3 + c = x^3 + c \]
c) \( \displaystyle \int \frac{3}{x} \, dx \)
Solution for (c):
Here \( n = -1 \). Using the formula \( \int \frac{a}{x} \, dx = a \ln |x| + c \):
\[ \int \frac{3}{x} \, dx = \int 3x^{-1} \, dx = 3 \ln |x| + c \]
d) \( \displaystyle \int \sqrt{x} \, dx \)
Solution for (d):
First rewrite: \( \sqrt{x} = x^{1/2} \), so \( n = \frac{1}{2} \).
\[ \int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{1}{\frac{1}{2}+1} x^{\frac{1}{2}+1} + c = \frac{1}{\frac{3}{2}} x^{3/2} + c = \frac{2}{3} x^{3/2} + c \]
This can also be written as \( \frac{2}{3} x \sqrt{x} + c \).
e) \( \displaystyle \int 5 \sqrt[3]{x^2} \, dx \)
Solution for (e):
First rewrite: \( \sqrt[3]{x^2} = x^{2/3} \), so \( a = 5, n = \frac{2}{3} \).
\[ \int 5 \sqrt[3]{x^2} \, dx = \int 5 x^{2/3} \, dx = \frac{5}{\frac{2}{3}+1} x^{\frac{2}{3}+1} + c = \frac{5}{\frac{5}{3}} x^{5/3} + c = 3 x^{5/3} + c \]
This can also be written as \( 3 x \sqrt[3]{x^2} + c \).
f) \( \displaystyle \int x^2 \sqrt[3]{x^2} \, dx \)
Solution for (f):
First rewrite: \( x^2 \sqrt[3]{x^2} = x^2 \cdot x^{2/3} = x^{2 + \frac{2}{3}} = x^{8/3} \), so \( n = \frac{8}{3} \).
\[ \int x^2 \sqrt[3]{x^2} \, dx = \int x^{8/3} \, dx = \frac{1}{\frac{8}{3}+1} x^{\frac{8}{3}+1} + c = \frac{1}{\frac{11}{3}} x^{11/3} + c = \frac{3}{11} x^{11/3} + c \]
This can also be written as \( \frac{3}{11} x^{3} \sqrt[3]{x^2} + c \).
Properties of Indefinite Integrals
To simplify integration, master these properties:
1. \( \displaystyle \int k \, dx = kx + c \), where \( k \) is a constant.
2. \( \displaystyle \int k f(x) \, dx = k \int f(x) \, dx \) (constant can be factored out).
3. \( \displaystyle \int [f(x) + g(x)] \, dx = \int f(x) \, dx + \int g(x) \, dx \)
4. \( \displaystyle \int [f(x) - g(x)] \, dx = \int f(x) \, dx - \int g(x) \, dx \)
Note: For properties (3) and (4), if a function has several terms, each term can be integrated separately. For products or quotients of functions, expand or simplify first into a sum of power functions \( a x^n \), then integrate term by term.
Example 2: Applying Properties
Find the following integrals:
a) \( \displaystyle \int 3 \, dx \)
Solution for (a):
Using Property 1:
\[ \int 3 \, dx = 3x + c \]
b) \( \displaystyle \int 3x^5 \, dx \)
Solution for (b):
Using Property 2:
\[ \int 3x^5 \, dx = 3 \int x^5 \, dx = 3 \cdot \frac{1}{6} x^6 + c = \frac{1}{2} x^6 + c \]
c) \( \displaystyle \int (x^2 + x) \, dx \)
Solution for (c):
Using Property 3:
\[ \int (x^2 + x) \, dx = \int x^2 \, dx + \int x \, dx = \frac{1}{3} x^3 + \frac{1}{2} x^2 + c \]
d) \( \displaystyle \int (x^2 - x) \, dx \)
Solution for (d):
Using Property 4:
\[ \int (x^2 - x) \, dx = \int x^2 \, dx - \int x \, dx = \frac{1}{3} x^3 - \frac{1}{2} x^2 + c \]
e) \( \displaystyle \int (x^3 - 2x + 5) \, dx \)
Solution for (e):
Integrate each term separately:
\[ \int (x^3 - 2x + 5) \, dx = \int x^3 \, dx - \int 2x \, dx + \int 5 \, dx = \frac{1}{4} x^4 - x^2 + 5x + c \]
f) \( \displaystyle \int (x^2 + 2)(2x - 3) \, dx \)
Solution for (f):
First expand the product:
\[ (x^2 + 2)(2x - 3) = 2x^3 - 3x^2 + 4x - 6 \]
Now integrate term by term:
\[ \int (2x^3 - 3x^2 + 4x - 6) \, dx = \frac{1}{2} x^4 - x^3 + 2x^2 - 6x + c \]
g) \( \displaystyle \int \frac{x^4 + 2x^2 - 1}{3x^2} \, dx \)
Solution for (g):
First simplify the fraction:
\[ \frac{x^4 + 2x^2 - 1}{3x^2} = \frac{x^2}{3} + \frac{2}{3} - \frac{1}{3x^2} = \frac{1}{3}x^2 + \frac{2}{3} - \frac{1}{3} x^{-2} \]
Now integrate term by term:
\[ \int \left( \frac{1}{3}x^2 + \frac{2}{3} - \frac{1}{3} x^{-2} \right) dx = \frac{1}{9} x^3 + \frac{2}{3}x + \frac{1}{3x} + c \]
h) \( \displaystyle \int \frac{x + 4}{\sqrt{x}} \, dx \)
Solution for (h):
Rewrite the integrand:
\[ \frac{x + 4}{\sqrt{x}} = \frac{x}{\sqrt{x}} + \frac{4}{\sqrt{x}} = x^{1/2} + 4x^{-1/2} \]
Now integrate term by term:
\[ \int \left( x^{1/2} + 4x^{-1/2} \right) dx = \frac{2}{3} x^{3/2} + 8 x^{1/2} + c = \frac{2}{3} x \sqrt{x} + 8 \sqrt{x} + c \]
i) \( \displaystyle \int \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 \, dx \)
Solution for (i):
First expand the square:
\[ \left( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 = x - 2 + \frac{1}{x} \]
Now integrate term by term:
\[ \int \left( x - 2 + \frac{1}{x} \right) dx = \frac{1}{2} x^2 - 2x + \ln |x| + c \]
Proof of the Algebraic Integration Formulas
Recall the basic derivative rules: \( \frac{d}{dx} (x^n) = n x^{n-1} \) and \( \frac{d}{dx} (\ln x) = \frac{1}{x} \).
According to the definition of the integral, \( \int f(x) \, dx = F(x) + c \) is correct if \( \frac{d}{dx} (F(x) + c) = f(x) \).
Proof of \( \int x^n \, dx = \frac{1}{n+1} x^{n+1} + c \):
\( \displaystyle \frac{d}{dx} \left( \frac{1}{n+1} x^{n+1} + c \right) = (n+1) \cdot \frac{1}{n+1} x^{(n+1)-1} = x^n \) ✓
Proof of \( \int a x^n \, dx = \frac{a}{n+1} x^{n+1} + c \):
\( \displaystyle \frac{d}{dx} \left( \frac{a}{n+1} x^{n+1} + c \right) = (n+1) \cdot \frac{a}{n+1} x^{(n+1)-1} = a x^n \) ✓
Proof of \( \int \frac{1}{x} \, dx = \ln |x| + c \):
\( \displaystyle \frac{d}{dx} (\ln |x| + c) = \frac{1}{x} \) ✓
Proof of \( \int \frac{a}{x} \, dx = a \ln |x| + c \):
\( \displaystyle \frac{d}{dx} (a \ln |x| + c) = a \cdot \frac{1}{x} = \frac{a}{x} \) ✓