Properties of Indefinite Integral

Power Rule

This is one of the most basic and important rules in integration. If you encounter an integral in power form, use this formula:

\int x^n \, dx = \frac{1}{n+1}x^{n+1} + C, \quad \text{where } n \neq -1

Simply put, add one to the exponent, then divide by that new exponent.

For example, let's solve $\int x^5 \, dx$ .

Here, $n=5$ , so:

\int x^5 \, dx = \frac{1}{5+1}x^{5+1} + C

= \frac{1}{6}x^6 + C

Constant Multiple Rule

If the function to be integrated has a coefficient or constant, you can "pull" that constant out of the integral to simplify the calculation.

\int a \cdot f(x) \, dx = a \int f(x) \, dx

Remember, a is a constant number. This property is very useful for simplifying an integral before solving it.

Sum and Difference Rule

What if we have to integrate two functions that are added or subtracted? Easy, we can separate them into two different integrals.

\int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \pm \int g(x) \, dx

For example, to solve $\int (x^4 - x^3) \, dx$ , we first separate them:

\int (x^4 - x^3) \, dx = \int x^4 \, dx - \int x^3 \, dx

= \left( \frac{1}{4+1}x^5 \right) - \left( \frac{1}{3+1}x^4 \right) + C

= \frac{1}{5}x^5 - \frac{1}{4}x^4 + C

Don't forget to add only one constant C at the end. Technically, each integral would produce its own constant (e.g., $C_1$ and $C_2$ ). However, since all these constants are of an indeterminate value, their sum or difference will also result in a new indeterminate constant. Therefore, we just write it as a single C at the end.

Substitution Rule

For integrals that look complicated, like the product of two functions where one is the derivative of the other (or a multiple of it), we can use the substitution rule. The basic idea is to simplify the integral by replacing part of the function with a new variable, usually u.

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du

This is often called u-substitution. This method is very powerful for solving complex integrals.

Exercises

Find the result of $\int 6x^2 \, dx$ !
Find the result of $\int (3x^2 + 2x - 5) \, dx$ !
Solve the integral $\int 2x(x^2 + 3)^4 \, dx$ using the substitution rule!

Answer Key

To solve $\int 6x^2 \, dx$ , we can use the Constant Multiple Rule and the Power Rule.

Step 1: Pull the constant 6 out of the integral.

$\int 6x^2 \, dx = 6 \int x^2 \, dx$

Step 2: Use the power rule on $\int x^2 \, dx$ , where $n=2$ .

$6 \left( \frac{1}{2+1}x^{2+1} \right) + C$
$= 6 \left( \frac{1}{3}x^3 \right) + C$

Step 3: Multiply the constants to get the final result.

$= 2x^3 + C$
For the integral $\int (3x^2 + 2x - 5) \, dx$ , we use the Sum and Difference Rule to break it down into three separate integrals.

Step 1: Separate each term into its own integral.

$\int 3x^2 \, dx + \int 2x \, dx - \int 5 \, dx$

Step 2: Solve each integral one by one using the power and constant rules.

$= 3\left(\frac{1}{3}x^3\right) + 2\left(\frac{1}{2}x^2\right) - 5x + C$
$= x^3 + x^2 - 5x + C$

So, the answer is $x^3 + x^2 - 5x + C$ .
The integral $\int 2x(x^2 + 3)^4 \, dx$ is a classic example for the Substitution Rule.

Step 1: Choose a part of the function to be $u$ . A good choice is the part inside the parentheses.

Let: $u = x^2 + 3$ .

Step 2: Find the derivative of $u$ with respect to $x$ , which is $du/dx$ .

$\frac{du}{dx} = 2x$

From this, we can write $du = 2x \, dx$ .

Step 3: Perform the substitution. Replace $x^2+3$ with $u$ and $2x \, dx$ with $du$ .

$\int (x^2 + 3)^4 (2x \, dx) = \int u^4 \, du$

Step 4: Solve the simplified integral using the power rule.

$\int u^4 \, du = \frac{1}{5}u^5 + C$

Step 5: Substitute $u$ back to its original form.

$= \frac{1}{5}(x^2 + 3)^5 + C$

This is the final result.