Two Antiderivatives Revision Notes for VCE SSCE Mathematical Methods

Two Antiderivatives

When we integrate expressions involving linear functions raised to a power, we encounter two distinct cases that require different approaches. Understanding these two antiderivatives is essential for solving a wide range of integration problems.

The power rule for $(ax+b)^r$ when $r \neq -1$

When integrating a linear expression raised to any power except $-1$ , we can use a straightforward formula derived from reversing the chain rule.

Consider a function of the form $(ax+b)^{r+1}$ where $r \neq -1$ . Using the chain rule to differentiate this gives us:

f'(x) = a(r+1)(ax+b)^r

Therefore, when we integrate $(ax+b)^r$ , we must reverse this process:

\int (ax+b)^r \, dx = \frac{1}{a(r+1)}(ax+b)^{r+1} + c, \quad r \neq -1

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Key points to remember:

Increase the power by $1$ (from $r$ to $r+1$ )
Divide by the coefficient of $x$ (which is $a$ ) multiplied by the new power
Don't forget the constant of integration $c$
This formula does NOT work when $r = -1$

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Worked Example: Power Functions

Let's find the general antiderivative of $(3x+1)^5$ .

\int (3x+1)^5 \, dx = \frac{1}{3(5+1)}(3x+1)^6 + c

= \frac{1}{18}(3x+1)^6 + c

Now consider $(2x-1)^{-2}$ :

\int (2x-1)^{-2} \, dx = \frac{1}{2(-2+1)}(2x-1)^{-1} + c

= -\frac{1}{2}(2x-1)^{-1} + c

This can also be written as:

= -\frac{1}{2(2x-1)} + c

The logarithmic case when $r = -1$

The special case occurs when $r = -1$ , which means we're trying to integrate $\frac{1}{ax+b}$ . The power rule cannot be used here because it would involve division by zero.

Instead, we need to recall that the derivative of $\log_e x$ is $\frac{1}{x}$ . This connection to logarithms is what makes $r = -1$ a special case.

Understanding the positive and negative cases

The antiderivative of $\frac{1}{ax+b}$ depends on whether $ax+b$ is positive or negative:

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For $ax + b > 0$ :

\int \frac{1}{ax+b} \, dx = \frac{1}{a}\log_e(ax+b) + c

For $ax + b < 0$ :

We need to use $\log_e(-ax-b)$ instead. This is because when $x < 0$ , the derivative of $\log_e(-x)$ equals $\frac{1}{x}$ :

\frac{d}{dx}\left[\log_e(-x)\right] = \frac{1}{-x} \times (-1) = \frac{1}{x}

Therefore:

\int \frac{1}{ax+b} \, dx = \frac{1}{a}\log_e(-ax-b) + c

Simplifying with absolute value

To avoid dealing with separate cases, we can use the absolute value function. The absolute value is defined as:

|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

For example, $|-2| = 2$ and $|2| = 2$ .

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Using absolute value notation, we can write a single formula that works for all cases where $ax+b \neq 0$ :

\int \frac{1}{ax+b} \, dx = \frac{1}{a}\log_e|ax+b| + c

This approach is recommended because it avoids confusion and is consistent with how calculators display the result.

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Worked Example: Logarithmic Integration

Find the general antiderivative of $\frac{2}{3x-2}$ for $x > \frac{2}{3}$ .

Since $x > \frac{2}{3}$ , we know that $3x - 2 > 0$ , so:

\int \frac{2}{3x-2} \, dx = \frac{1}{3} \times 2\log_e(3x-2) + c

= \frac{2}{3}\log_e(3x-2) + c

For $x < \frac{2}{3}$ , we would have $3x - 2 < 0$ , giving:

\int \frac{2}{3x-2} \, dx = \frac{2}{3}\log_e(2-3x) + c

Using absolute value notation, we can write both cases as:

\int \frac{2}{3x-2} \, dx = \frac{2}{3}\log_e|3x-2| + c

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Worked Example: Finding a Particular Solution

Given $\frac{dy}{dx} = \frac{3}{x}$ for $x > 0$ and $y = 10$ when $x = 1$ , find an expression for $y$ in terms of $x$ .

First, integrate to find the general solution:

y = \int \frac{3}{x} \, dx = 3\log_e x + c

Now use the initial condition $y = 10$ when $x = 1$ :

10 = 3\log_e 1 + c

10 = 0 + c

c = 10

Therefore:

y = 3\log_e x + 10

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For $x < 0$ with the same derivative but $y = 10$ when $x = -1$ :

y = \int \frac{3}{x} \, dx = 3\log_e(-x) + c

10 = 3\log_e 1 + c

c = 10

y = 3\log_e(-x) + 10

Notice that using absolute value notation, both solutions can be written as $y = 3\log_e|x| + 10$ .

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Worked Example: Another Logarithmic Case

Find the general antiderivative of $\frac{2}{2-3x}$ .

\int \frac{2}{2-3x} \, dx = -\frac{1}{3} \times 2\log_e|2-3x| + c

= -\frac{2}{3}\log_e|2-3x| + c

Note the negative sign appears because the coefficient of $x$ in the denominator is $-3$ .

Summary of the two cases

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Case 1: When $r \neq -1$ :

\int (ax+b)^r \, dx = \frac{1}{a(r+1)}(ax+b)^{r+1} + c

Case 2: When $r = -1$ (and $ax+b \neq 0$ ):

\int \frac{1}{ax+b} \, dx = \frac{1}{a}\log_e|ax+b| + c

Remember!

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Essential Points to Remember:

The power rule for $(ax+b)^r$ works for all values of $r$ except $r = -1$
When $r = -1$ , the antiderivative involves a natural logarithm, not a power function
Always use absolute value notation in logarithmic antiderivatives: $\log_e|ax+b|$
Don't forget to divide by the coefficient $a$ in both cases
The constant of integration $c$ must be included for general antiderivatives

Two Antiderivatives (VCE SSCE Mathematical Methods): Revision Notes