## DefinitionEdit

Now recall that *F* is said to be an antiderivative of *f* if . However, *F* is not the only antiderivative. We can add any constant to *F* without changing the derivative. With this, we define the **indefinite integral** as follows:

*F*satisfies and

*C*is any constant.

The function , the function being integrated, is known as the **integrand**. Note that the indefinite integral yields a *family* of functions.

**Example**

Since the derivative of is , the general antiderivative of is plus a constant. Thus,

**Example: Finding antiderivatives**

Let's take a look at . How would we go about finding the integral of this function? Recall the rule from differentiation that

In our circumstance, we have:

This is a start! We now know that the function we seek will have a power of 3 in it. How would we get the constant of 6? Well,

Thus, we say that is an antiderivative of .

### ExercisesEdit

## Indefinite integral identitiesEdit

### Basic Properties of Indefinite IntegralsEdit

**Constant Rule for indefinite integrals**

*c*is a constant then

**Sum/Difference Rule for indefinite integrals**

### Indefinite integrals of PolynomialsEdit

Say we are given a function of the form, , and would like to determine the antiderivative of *f*. Considering that

we have the following rule for indefinite integrals:

**Power rule for indefinite integrals**

### Integral of the Inverse functionEdit

To integrate , we should first remember

Therefore, since is the derivative of we can conclude that

Note that the polynomial integration rule does not apply when the exponent is -1. This technique of integration must be used instead. Since the argument of the natural logarithm function must be positive (on the real line), the absolute value signs are added around its argument to ensure that the argument is positive.

### Integral of the Exponential functionEdit

Since

we see that is its own antiderivative. This allows us to find the integral of an exponential function:

### Integral of Sine and CosineEdit

Recall that

So *sin x* is an antiderivative of *cos x* and *-cos x* is an antiderivative of *sin x*. Hence we get the following rules for integrating *sin x* and *cos x*

We will find how to integrate more complicated trigonometric functions in the chapter on integration techniques.

**Example**

Suppose we want to integrate the function . An application of the sum rule from above allows us to use the power rule and our rule for integrating as follows,

### ExercisesEdit

## The Substitution RuleEdit

The substitution rule is a valuable asset in the toolbox of any integration greasemonkey. It is essentially the chain rule (a differentiation technique you should be familiar with) in reverse. First, let's take a look at an example:

### Preliminary ExampleEdit

Suppose we want to find . That is, we want to find a function such that its derivative equals . Stated yet another way, we want to find an antiderivative of . Since differentiates to , as a first guess we might try the function . But by the Chain Rule,

Which is almost what we want apart from the fact that there is an extra factor of 2 in front. But this is easily dealt with because we can divide by a constant (in this case 2). So,

Thus, we have discovered a function, , whose derivative is . That is, *F* is an antiderivative of . This gives us

### GeneralizationEdit

In fact, this technique will work for more general integrands. Suppose *u* is a differentiable function. Then to evaluate we just have to notice that by the Chain Rule

As long as is continuous we have that

Now the right hand side of this equation is just the integral of but with respect to *u*. If we write *u* instead of *u(x)* this becomes

So, for instance, if we have worked out that

### General Substitution RuleEdit

Now there was nothing special about using the cosine function in the discussion above, and it could be replaced by any other function. Doing this gives us the substitution rule for indefinite integrals:

**Substitution rule for indefinite integrals**

Assume *u* is differentiable with continuous derivative and that *f* is continuous on the range of *u*. Then

Notice that it looks like you can "cancel" in the expression to leave just a . This does not really make any sense because is **not a fraction**. But it's a good way to remember the substitution rule.

### ExamplesEdit

The following example shows how powerful a technique substitution can be. At first glance the following integral seems intractable, but after a little simplification, it's possible to tackle using substitution.

**Example**

We will show that

First, we re-write the integral:

- .

Now we preform the following substitution:

Which yields:

- .

### ExercisesEdit

## Integration by PartsEdit

Integration by parts is another powerful tool for integration. It was mentioned above that one could consider integration by substitution as an application of the chain rule in reverse. In a similar manner, one may consider integration by parts as the product rule in reverse.

### Preliminary ExampleEdit

### General Integration by PartsEdit

**Integration by parts for indefinite integrals**

Suppose *f* and *g* are differentiable and their derivatives are continuous. Then

If we write *u=f(x)* and *v=g(x)*, then by using the Leibniz notation *du=f'(x) dx* and *dv=g'(x) dx* the integration by parts rule becomes

### ExamplesEdit

**Example**

Find

Here we let:

- , so that ,
- , so that .

Then:

**Example**

Find

In this example we will have to use integration by parts twice.

Here we let

- , so that ,
- , so that .

Then:

Now to calculate the last integral we use integration by parts again. Let

- , so that ,
- , so that

and integrating by parts gives

So, finally we obtain

**Example**

Find

The trick here is to write this integral as

Now let

- so ,
- so .

Then using integration by parts,

**Example**

Find

Again the trick here is to write the integrand as . Then let

*u*= arctan(*x*); d*u*= 1/(1+*x*^{2}) d*x**v*=*x*; d*v*= 1·d*x*

so using integration by parts,

**Example**

Find

This example uses integration by parts twice. First let,

*u*= e^{x}; thus d*u*= e^{x}d*x*- d
*v*= cos(*x*)d*x*; thus*v*= sin(*x*)

so

Now, to evaluate the remaining integral, we use integration by parts again, with

*u*= e^{x}; d*u*= e^{x}d*x**v*= -cos(*x*); d*v*= sin(*x*)d*x*

Then

Putting these together, we have

Notice that the same integral shows up on both sides of this equation, but with opposite signs. The integral does not cancel; it doubles when we add the integral to both sides to get

### ExercisesEdit