Calculus/Indefinite integral
Contents
Definition[edit]
Now recall that is said to be an antiderivative of f if . However, is not the only antiderivative. We can add any constant to without changing the derivative. With this, we define the indefinite integral as follows:
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 .
Exercises[edit]
Indefinite integral identities[edit]
Basic Properties of Indefinite Integrals[edit]
Constant Rule for indefinite integrals
Sum/Difference Rule for indefinite integrals
Indefinite integrals of Polynomials[edit]
Say we are given a function of the form, , and would like to determine the antiderivative of . Considering that
we have the following rule for indefinite integrals:
Power rule for indefinite integrals
for allIntegral of the Inverse function[edit]
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 . 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 function[edit]
Since
we see that is its own antiderivative. This allows us to find the integral of an exponential function:
Integral of Sine and Cosine[edit]
Recall that
So is an antiderivative of and is an antiderivative of . Hence we get the following rules for integrating and
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,

.
Exercises[edit]
The Substitution Rule[edit]
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 Example[edit]
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, is an antiderivative of . This gives us
Generalization[edit]
In fact, this technique will work for more general integrands. Suppose 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 . If we write instead of this becomes
So, for instance, if we have worked out that
General Substitution Rule[edit]
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 is differentiable with continuous derivative and that is continuous on the range of . 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.
Examples[edit]
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 rewrite the integral:
Now we perform the following substitution:
 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle \frac{du}{dx}=2a^2x^{3}\ \implies\ x^{3}dx=\frac{du}{2a^2}}
Which yields:

 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle =\frac{1}{2a^2}\left(\frac{2}{\sqrt u}\right)+C}
 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle =\frac{x}{a^2 \sqrt{x^2+a^2}}+C}
Exercises[edit]
Integration by Parts[edit]
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 Example[edit]
General Integration by Parts[edit]
Integration by parts for indefinite integrals
Suppose and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): g
are differentiable and their derivatives are continuous. Then
it is also very important to notice that
to set the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): f(x)
and we need to follow the rule called I.L.A.T.E.
ILATE defines the order in which we must set the
 I for inverse trigonometric function
 L for log functions
 A for algebraic functions
 T for trigonometric functions
 E for exponential function
f(x) and g(x) must be in the order of ILATE or else your final answers will not match with the main key
Examples[edit]
Example
Find Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle \int x\cos(x)dx}
Here we let:
 , so that ,
 , so that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle v=\sin(x)} .
Then:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): {\displaystyle =uv\int v\,du}
Example
Find
In this example we will have to use integration by parts twice.
Here we let
 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): u=x^{2} , so that ,
 , so that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "/mathoid/local/v1/":): v=e^{x} .
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
 so
 so
so using integration by parts,
Example
Find
This example uses integration by parts twice. First let,
 so
 so
so
Now, to evaluate the remaining integral, we use integration by parts again, with
 so
 so
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
Exercises[edit]