Properties of Indefinite Integral

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Increase the exponent by one, then divide by the $new exponent$ . Apply the formula on an interval where the integrand is defined. For an arbitrary real exponent, we can use an interval with $x>0$ .

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 $can be moved outside an integral$ :

If $F'(x)=f(x)$ , multiplying an antiderivative by a constant gives:

\int a f(x)\,dx=aF(x)+C

Keep the constant of integration, including when $a=0$ . For example, the factor $-3$ in $-3x^2$ remains after integration:

\int -3x^2\,dx=-3\frac{x^3}{3}+C=-x^3+C

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

The first two problems use $linearity of integration$ and the power rule. In the third, the expression $x^2+3$ appears alongside its derivative $2x$ , suggesting substitution.

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

$Problem 1$ . 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 $n=2$ .

\int 6x^2\,dx=6\left(\frac{x^{2+1}}{2+1}\right)+C

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

\int 6x^2\,dx=2x^3+C

$Problem 2$ . 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+2x-5)\,dx=\int3x^2\,dx+\int2x\,dx-\int5\,dx

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

\int(3x^2+2x-5)\,dx=3\left(\frac{x^3}{3}\right)+2\left(\frac{x^2}{2}\right)-5x+C

So, the answer is $x^3 + x^2 - 5x + C$ .

$Problem 3$ . The integral $\int 2x(x^2 + 3)^4 \, dx$ allows $substitution$ because the factor $2x$ is the derivative of the expression in parentheses.

$Step 1$ : 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.

\int2x(x^2+3)^4\,dx=\frac15(x^2+3)^5+C

Check the answer by $differentiating it$ . The chain rule returns the original integrand:

\frac{d}{dx}\left[\frac15(x^2+3)^5+C\right]=\frac15\cdot5(x^2+3)^4\cdot2x=2x(x^2+3)^4