Properties of Indefinite Integral

Power Rule

For a power of the integration variable, use:

\begin{aligned} &\int x^n \, dx = \frac{1}{n+1}x^{n+1} + C, \\ &\text{where } n \neq -1 \end{aligned}

Increase the exponent by one, then divide by that new exponent.

The value $n=-1$ is excluded because it would make the denominator zero. That case follows the logarithmic rule $\int x^{-1}\,dx=\ln|x|+C$ .

For example, evaluate $\int x^5 \, dx$ .

Here, $n=5$ , so:

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

Constant Multiple Rule

A constant factor can be moved outside an integral:

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

The rule requires $a$ to be constant with respect to the integration variable.

Sum and Difference Rule

A sum or difference can be integrated term by term.

\begin{aligned} &\int [f(x) \pm g(x)] \, dx = \int f(x) \, dx \\ &\pm \int g(x) \, dx \end{aligned}

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

Write one constant $C$ at the end. Formally, we could use separate constants $C_1$ and $C_2$ for the two antiderivatives, but their sum or difference is again an arbitrary constant. A single $C$ therefore covers every possible value of that constant.

Substitution Rule

When an integrand contains a composite expression together with its derivative or a constant multiple of it, set the inner expression equal to a new variable, usually $u=g(x)$ . Then $du=g'(x)\,dx$ replaces the matching derivative factor.

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

This change of variable is called u-substitution. It turns the nested integrand into $f(u)$ when the matching derivative factor is present.

Exercises

Each problem uses the linearity rules, so split the sum into separate integrals, factor out the constant, and then integrate each term with the power rule.

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!

Worked Solutions

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 $n=2$ .
$6 \left( \frac{1}{2+1}x^{2+1} \right) + C$
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.
$= 3\left(\frac{1}{3}x^3\right) + 2\left(\frac{1}{2}x^2\right) - 5x + C$
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.