We start with the first rule of integration, which allows us to start calculating explicit integrals.
Definition
One of the most fundamental rules in differential calculus is the power rule,
\frac{d}{dx} x^m = mx^{m-1} .
Suspecting this will be a useful property, we can derive an expression for the inverse power rule from the definition of an integral.
\int mx^{m-1} dx = x^m
Applying linearity and the substitution n = m-1, we obtain
\int x^{n} dx = \frac{x^{n+1}}{n+1}
whenever n \not = -1. This is known as the inverse power rule, and allows us to integrate any regular polynomial.
For every n \in \mathbb{R} \backslash \{-1\}
\int x^{n} dx = \frac{x^{n+1}}{n+1} + C.
This is the inverse power rule.
Polynomials
Lets apply this rule to some polynomials.
Find the solution to the following integral
\int 2x^2 - 3x - 1 \, dx
We first begin by applying linearity
= 2 \int x^2 \,dx - 3\int x \, dx - \int 1x^0 \, dx
Then inverse power rule
= \frac{2x^3}{3} - \frac{3x^2}{2} - x + C
This process can be done for any standard polynomial.
Let f(x) be a polynomial in the form
f(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + c
Then
\int f(x)\,dx = \frac{a_nx^{n+1}}{n+1} + \frac{a_{n-1}x^{n}}{n} + \cdots + \frac{a_1x^2}{2} + cx + C
Edge case and roots
There begs the question though, how do we integrate when n = -1? The inverse power rule gives us an undefined answer
\int x^{-1}\,dx = \frac{1}{0} + C.
The solution, once again provided without proof, is to use the natural logarithm.
\int x^{-1}\,dx = \ln|x| + C.
Note the absolute value bars. This is because the function \ln is not defined for negative values of x, whereas our integrand is.
Since the inverse power rule works for roots, we’re able to take the integral of arbitrary root functions.
Find the solution to the following integral
\int \frac{x}{\sqrt[3]{x}} \, dx
Changing exponents allows us to get the following
\int \frac{x}{\sqrt[3]{x}} \, dx = \int x^{2/3} \, dx
Then inverse power rule
= \frac{3x^{5/3}}{5} + C
Examples
Evaluate the following integral
\int 6x^5\, dx - 18x^2 + 7
Notice the integral ends at the dx, meaning the solution is
\int 6x^5 - 18x^2 + 7 \, dx = x^6 - 18x^2 + C
Evaluate the following integral
\int 12t^7 - t^2 - t + 3 \, dt
Classic polynomial, the only difference is the use of t instead of x
\int 12t^7 - t^2 - t + 3 \, dt = \frac{3}{2}t^8 - \frac{1}{3}t^3 - \frac{1}{2}t^2 + 3t + C
Evaluate the following integral
\int \sqrt[3]{w} + 10\sqrt[5]{w^3}\, dw
Convert the radicals to fractional exponents
\int \sqrt[3]{w} + 10\sqrt[5]{w^3}\, dw = \int w^\frac{1}{3} + 10w^\frac{3}{5} \, dw
And then apply inverse power rule
\int \sqrt[3]{w} + 10\sqrt[5]{w^3}\, dw = \frac{3}{4}w^\frac{4}{3} + \frac{25}{4}w^\frac{8}{5} + C
Evaluate the following integral
\int \frac{7}{3y^6} + \frac{1}{y^{10}} - \frac{2}{\sqrt[3]{y^4}} \, dy
Convert the radicals to fractional exponents
\int \frac{7}{3y^6} + \frac{1}{y^{10}} - \frac{2}{\sqrt[3]{y^4}} \, dy = \int \frac{7}{3}y^{-6} + y^{-10} - 2y^{\frac{4}{3}} \, dy
And then apply inverse power rule
\int \frac{7}{3y^6} + \frac{1}{y^{10}} - \frac{2}{\sqrt[3]{y^4}} \, dy = -\frac{7}{15}y^{-5} - \frac{1}{9} y^{-9} + 6y^{-\frac{1}{3}} + C