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6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation

1 min readjune 18, 2024

6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation

Just like how there are basic rules for calculating derivatives, there are rules for calculating antiderivatives. Since antiderivatives are the inverses of derivatives, these rules are mostly the reverse of the basic derivative rules. 💡


📈 Indefinite Integrals: Notation

Let’s first talk about a family of functions before we dive into reversing the derivative process.

Imagine we have two different antiderivatives, F(x)=x2+3F(x) = x^2+3 and G(x)=x22G(x) = x^2-2.

If we were to take the derivative of both of these functions, we would find that they both have the same derivative, 2x2x. If we reverse the derivative process through integration, how do we account for arriving at these two different antiderivatives? Introducing the magical constant CC! 🪄

When we integrate 2x2x, the antiderivative is 2x+C2x+C where CC is any constant. This result is often referred to as a family of functions because they vary only in the value of their constants and all share the same derivative.

This type of integral is referred to as an indefinite integral because we can’t be sure which member of the family of antiderivatives is at play. If the bounds of the integral are not specified as they are in a definite integral, always add ‘+C’ to the end of your antiderivative!

Here’s a general look at the notation!

f(x)dx=F(x)+C\int f(x)dx=F(x)+C

Where F(x)=f(x)F'(x)=f(x) and CC represents the integration constant.


📏 Indefinite Integrals: Basic Rules

Now let’s look at how to reverse the process of some of the derivatives we learned early in our study of calculus.

Reverse Power Rule

First up, we have the reverse power rule. This essentially refers to how to take the indefinite integral of a function, which is the reverse of the power rule used for differentiation. Suppose we have the following function:

f(x)=xn+1n+1+Cf(x) = \frac{x^{n+1}}{n+1}+C

Where n1n \neq -1 since n=1n=-1 causes f(x)f(x) to be undefined.

What is its derivative?

If we recall the power rule for derivatives, we see that the derivative of f(x)f(x) is

f(x)=xnf'(x) = x^{n}

Now, what is the antiderivative of the derivative of f(x)f(x)?

Using the fact that antiderivatives and derivatives are inverses, we see that...

xndx=xn+1n+1+C\int x^{n}dx = \frac{x^{n+1}}{n+1}+C

This is the reverse power rule. You’re basically adding one to the exponent of each term and dividing by the new exponent!

Reverse Power Rule Example 1

Evaluate the following integral:

x3dx\int x^3dx

Using the reverse power rule, we see that

x3dx=x3+13+1+C=x44+C\int x^3dx=\frac{x^{3+1}}{3+1}+C=\frac{x^{4}}{4}+C

Reverse Power Rule Example 2

Give the following a try! A useful tip is to rewrite fractions with negative exponents. You can also apply this logic to radical functions, since they can be rewritten with fractional exponents.

(1x27x3+2x2x+4)dx\int(\frac{1}{x^2}-7x^3+2x^2-x+4)\, dx

When we rewrite the first term, we see that…

(x27x3+2x2x+4)dx\int(x^{-2}-7x^3+2x^2-x+4)\, dx

Once we use the reverse power rule and evaluate this integral term by term, we get:

(1x27x3+2x2x+4)dx=1x7x44+2x33x22+4x+C\int(\frac{1}{x^2}-7x^3+2x^2-x+4)\, dx = -\frac{1}{x} -\frac{7x^4}{4}+\frac{2x^3}{3}-\frac{x^2}{2}+4x+C

Sums and Multiples Rules for Antiderivatives

If you recall, we learned in Unit 2 that there were the sums and multiples rules for derivatives. Similarly, there are the sums and multiples rules for antiderivatives.

The sums rule states that

[f(x)+g(x)]dx=f(x)dx+g(x)dx\int \left[f(x)+g(x)\right]dx=\int f(x)dx + \int g(x)dx

The multiples rule states that

cf(x)dx=cf(x)dx\int c \cdot f(x)dx=c \int f(x)dx

Sums and Multiples Examples

Here are examples of these two rules in action, the first covering the sums rule and the second covering the multiples rule.

[x4+x2]dx=x4dx+x2dx\int \left[x^4+x^2\right]dx=\int x^4dx + \int x^2dx
5x6dx=5x6dx\int 5x^{6}dx=5\int x^6dx

Antiderivatives of Trigonometric Functions

When you’re first learning your trig antiderivatives, you may find it useful to think to yourself, “What has a derivative of…?”

Antiderivative of sin(x)\sin(x)

If you recall, ddx[cos(x)]=sin(x)\frac{d}{dx}[\cos(x)]=-\sin(x). This means that ddx[cos(x)]=sin(x)\frac{d}{dx}[-\cos(x)]=\sin(x). Therefore,

sin(x)dx=cos(x)+C\int \sin(x)dx=-\cos(x)+C

Antiderivative of cos(x)\cos(x)

If you recall, ddx[sin(x)]=cos(x)\frac{d}{dx}[\sin(x)]=\cos(x). Therefore,

cos(x)dx=sin(x)+C\int \cos(x)dx=\sin(x)+C

Other Antiderivatives of Trig Functions

I would also know the following trig integrals for the AP exam:

sec2(x)dx=tan(x)+C\int sec^2(x) \, dx = tan(x) +C
csc2(x)dx=cot(x)+C\int csc^2(x) \, dx = -cot(x) +C
sec(x)tan(x)dx=sec(x)+C\int sec(x)tan(x) \, dx = sec(x) +C
csc(x)cot(x)dx=csc(x)+C\int csc(x)cot(x) \, dx = -csc(x) +C

Antiderivatives of Inverse Trig Functions

These integrals aren’t nearly as common on the AP test, but below are the forms you may encounter on the AP test.

11x2dx=sin1(x)+C\int \frac{1}{\sqrt{1-x^2}} \, dx = sin^{-1}(x) +C
11+x2dx=tan1(x)+C\int \frac{1}{{1+x^2}} \, dx = tan^{-1}(x) +C

Antiderivatives of Transcendental Functions (1x\frac{1}{x}, exe^x)

Finally, we have the integrals for the transcendental functions you are likely to encounter on the AP exams.

Antiderivative of 1x\frac{1}{x}

If you recall, ddx[ln(x)]=1x\frac{d}{dx}[\ln(x)]=\frac{1}{x}. Therefore, it is not a bad guess to say that 1xdx=ln(x)+C\int \frac{1}{x}dx=\ln(x)+C.

However, because of the domain of ln(x)\ln(x), which is (0,)(0, \infty), if we want to be able to take the antiderivative of 1x\frac{1}{x} for any positive or negative xx, we need to rewrite this rule as

1xdx=lnx+C\int \frac{1}{x}dx=\ln\mid x \mid + C

Antiderivative of exe^{x}

If you recall, ddx[ex]=ex\frac{d}{dx}[e^{x}]=e^{x}. Therefore,

exdx=ex+C\int e^xdx=e^x+C

📝 Indefinite Integrals Practice Problems

Now that you know all the basic rules for antiderivatives, let’s do some practice problems!

❓ Indefinite Integrals Problems

Evaluate each of the following integrals.

1.x7dx=?1. \int x^7dx=?
2.[x4+cos(x)]dx=?2.\int \left[x^4+\cos(x)\right]dx=?
3.[4cos(x)+ex]dx=?3.\int \left[4\cos(x)+e^x\right]dx=?
4.(3x+x2)dx4.\int (\frac{3}{x}+x^2) \, dx

✅ Indefinite Integrals Question Solutions

Indefinite Integrals Question 1

When we take a look at question 1, we can quickly tell that we have to use the reverse power rule!

x7dx=x7+17+1+C=18x8+C\int x^7dx=\frac{x^{7+1}}{7+1}+C=\boxed{\frac{1}{8}x^{8}+C}

Indefinite Integrals Question 2

Using the sums rule for antiderivatives, we see that

[x4+cos(x)]dx=x4dx+cos(x)dx\int \left[x^4+\cos(x)\right]dx=\int x^4dx+\int \cos(x)dx

This means that we can take the antiderivatives of the two terms separately and then sum their individual antiderivatives together afterward.

Using the reverse power rule, we see that

x4dx=x4+14+1+C=15x5+C\int x^4dx = \frac{x^{4+1}}{4+1}+C=\frac{1}{5}x^5 + C

And using the antiderivative of cos(x)\cos(x), we see that

cos(x)dx=sin(x)+C\int \cos(x)dx = \sin(x)+C

Combining these two parts, we get

[x4+cos(x)]dx=15x5+sin(x)+C\int \left[x^4+\cos(x)\right]dx=\boxed{\frac{1}{5}x^5 +\sin(x)+ C}

Indefinite Integrals Question 3

Using the sums rule for antiderivatives, we see that

[4cos(x)+ex]dx=4cos(x)dx+exdx\int \left[4\cos(x)+e^x\right]dx=\int 4\cos(x)dx + \int e^xdx

This means that we can take the antiderivatives of the two terms separately and then sum their individual antiderivatives together afterward.

Using the multiples rule and the antiderivative of cos(x)\cos(x), we see that

4cos(x)dx=4cos(x)dx=4sin(x)+C\int 4\cos(x)dx=4\int \cos(x)dx=4\sin(x)+C

And using the antiderivative of exe^x, we see that

exdx=ex+C\int e^xdx=e^x+C

Combining these two parts, we get

[4cos(x)+ex]dx=4sin(x)+ex+C\int \left[4\cos(x)+e^x\right]dx=\boxed{4\sin(x)+e^x+C}

Indefinite Integrals Question 4

Using the sums rule for antiderivatives, we see that

(3x+x2)dx=3xdx+x2dx\int (\frac{3}{x}+x^2) \, dx=\int \frac{3}{x}dx+ \int x^2 \, dx

We can again take the antiderivatives of the two terms separately and then sum their individual antiderivatives together afterward.

Let’s take the integral of the first term, using the following rule 1xdx=lnx+C\int \frac{1}{x}dx=\ln\mid x \mid + C

(3x)dx=3ln(x)+C∫ (\frac{3}{x}) dx = 3 ln(|x|) + C

And using the reverse power rule, we see that

x2dx=x33+C∫ x^2 dx = \frac{x^3}{3} + C

Combining these two parts, we get

(3x+x2)dx=3ln(x)+x33+C\int (\frac{3}{x}+x^2) \, dx =\boxed{3ln(|x|) + \frac{x^3}{3}+C}

✨ Closing

Woah! We've covered the reverse power rule, sums and multiples rules for antiderivatives, antiderivatives of trigonometric functions, inverse trig functions, transcendental functions, and practiced.

My biggest tip? Remember that taking integrals involves the reverse process of differentiation and you must add +C+C to the end of your answer of an indefinite integral. Good luck! 🍀