Indefinite Integrals

Derivatives? Been there. Finding derivatives? Done that.

But how what about integrals? What about finding indefinite integrals?

Don’t worry — we’ve got you.

What is the indefinite integral?

Let $$f(x)$$ be a function. An antiderivative of $$f(x)$$ is any function $$F(x)$$ such that:

$$F'(x) = f(x)$$

Since there are a lot of antiderivatives of one function, the indefinite integral of f gives us all its antiderivatives:

$${\int f(x) dx = F(x) + C}$$

where $$C$$ is any constant, called the constant of integration.

Hmmm, okay, but how do we find this integral?

We use the properties and rules of integrals!

Properties of integrals are used to simplify the process of integration by splitting the procedure into several steps. Here’s a list of all the properties and rules you’ll want to keep handy:

 

Constant multiple of integrals $$\int{(c\times f(x))}dx=c\times \int{f(x)}dx$$
Sum rule for integrals $$\int{(f(x) + g(x))}dx=\int{f(x)}dx + \int{g(x)}dx$$
Sum rule for integrals $$\int{(f(x) - g(x))}dx=\int{f(x)}dx - \int{g(x)}dx$$
Substitution rule $$\int{f(\varphi(t))}\varphi^{\prime}(t)dt=\int{f(x)}dx$$
Integration by parts $$\int{u}dv=uv-\int{v}du$$

Why is the indefinite integral so useful?

Finding an indefinite integral is kind of “step one” for a lot of calculus, like in solving differential equations, or even in finding a definite integral!

In practice, we can use indefinite integrals to calculate displacement from velocity, velocity from acceleration, and so much more.

How to find the indefinite integral

Now that we know what the indefinite integral is and why it’s useful, it’s time to see it in action! Let’s walk through some examples together.

Example 1


Find the integral:

$$\int (t-1)t^5dt$$

Notice that the expression under the integral can be simplified, so distribute $$t^5$$ through the parentheses:

$$\int (t^6-t^5)dt$$

Use the property of integral $$\int f(x) \pm g(x)dx=\int f(x)dx\pm \int g(x)dx$$:

$$\int t^6dt-\int t^5dt$$

Now we have to use the integration rules. Here, we’ll use $$\int x^n dx=\frac{x^{n+1}}{n+1}, n \neq -1$$ to evaluate the integrals:

$$\frac{t^{6+1}}{6+1}-\frac{t^{5+1}}{5+1}$$

Add the numbers in the exponents and in the denominators:

$$\frac{t^{7}}{7}-\frac{t^{6}}{6}$$

Since the derivative of a constant is zero, we need to add the constant of integration $$C$$ to include all possible anti-derivative functions in the result:

$$\frac{t^{7}}{7}-\frac{t^{6}}{6}+C, C\in \mathbb{R}$$

So the indefinite integral $$\int (t-1)t^5dt$$ is equal to:

$$\frac{t^{7}}{7}-\frac{t^{6}}{6}+C, C\in \mathbb{R}$$

Not too bad, right? Let’s try one more.

Example 2


Find the integral:

$$\int\frac{(3x - 2)^{2}}{x^{3}}dx$$

Use $$(a-b)^2=a^2-2ab+b^2$$ to expand the expression in the numerator:

$$\int\frac{9x^{2}-12x+4}{x^{3}}dx$$

We have three terms in the numerator, so let’s separate the fraction into three fractions:

$$\int\frac{9x^{2}}{x^{3}}-\frac{12x}{x^{3}}+\frac{4}{x^{3}}dx$$

Notice that there’s $$x$$ in the numerator and denominator of the first two fractions, so we can simplify those:

$$\int\frac{9}{x}-\frac{12}{x^{2}}+\frac{4}{x^{3}}dx$$

Use the property of integral $$\int f(x) \pm g(x)dx=\int f(x)dx\pm \int g(x)dx$$:

$$\int\frac{9}{x}dx-\int\frac{12}{x^{2}}dx+\int\frac{4}{x^{3}}dx$$

Time to use the integration rules! We’ll use $$\int \frac{a}{x} dx=a\times \ln{(|x|)}$$ to evaluate the integral:

$$9\ln{(|x|)}-\int\frac{12}{x^{2}}dx+\int\frac{4}{x^{3}}dx$$

To evaluate the second integral, first use $$\int a\times f(x) dx=a\times\int f(x)dx$$:

$$9\ln{(|x|)}-12\times\int\frac{1}{x^{2}}dx+\int\frac{4}{x^{3}}dx$$

Now, use $$\int \frac{1}{x^n}dx=-\frac1{(n-1)\times x^{n-1}}$$ to evaluate the integral:

$$9\ln{(|x|)}-12\times(-\frac1x)+\int\frac{4}{x^{3}}dx$$

To evaluate the third integral, first use $$\int a\times f(x) dx=a\times\int f(x)dx$$:

$$9\ln{(|x|)}-12\times(-\frac1x)+4\times\int\frac{1}{x^{3}}dx$$

Now, use $$\int \frac{1}{x^n}dx=-\frac1{(n-1)\times x^{n-1}}$$ to evaluate the integral:

$$9\ln{(|x|)}-12\times(-\frac1x)+4\times(-\frac1{2x^2})$$

Simplify the expressions:

$$9\ln{(|x|)}+\frac{12}x-\frac{2}{x^2}$$

Since the derivative of a constant is zero, we need to add the constant of integration $$C$$ to include all possible anti-derivative functions in the result:

$$9\ln{(|x|)}+\frac{12}x-\frac{2}{x^2}+C, C\in \mathbb{R}$$

So, the indefinite integral $$\int\frac{(3x – 2)^{2}}{x^{3}}dx$$ is equal to:

$$9\ln{(|x|)}+\frac{12}x-\frac{2}{x^2}+C, C\in \mathbb{R}$$

Great job!

You can apply that process to whatever problem you choose. Just keep these steps in mind:

Study summary

  1. Simplify the expression, if possible.
  2. Use the properties of an integral.
  3. Evaluate the integral.
  4. Add the constant of integration.

Do it yourself!

Even if you don’t want to, you’ll feel better once you get a few practice problems under your belt. Going through the steps of each one helps you understand the process so much better! Try these examples and see how you do:

Find the integral:

1.$$\int\frac{5x+1}{x^{2}}dx$$

2. $$\int(3x^2-\ln x)dx$$
3. $$\int(\cos(t^2)+\frac{1}{t})dt$$
4. $$\int(e^x-e^{4x})dx$$

Solutions:

1. $$5\ln{(|x|)}-\frac1x+C, C \in \mathbb{R}$$
2. $$x^3-\ln{(x)}\times x +x+C, C \in \mathbb{R}$$
3. $$\text{No elementary antiderivative}$$
4. $$e^x-\frac{e^{4x}}{4}+C, C \in \mathbb{R}$$

Need help? We’re here! Just scan the problem with your Photomath app and we’ll help you walk through each step.

Here’s a sneak peek of what you’ll see:

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