1. Antiderivatives for exponential functions
Recall that for f (x) = ec⋅x , f ′ (x) = c ⋅ ec⋅x (for any constant c). That is, ex is its
own derivative. So it makes sense that it is its own antiderivative as well!
Theorem 1.1 (Antiderivatives of exponential functions). Let f (x) = ec⋅x for some
1
constant c. Then F (x) = ec⋅c + D, for any constant D, is an antiderivative of
c
f (x).
Proof. Consider F (x) = 1c ec⋅x + D. Then by the chain rule, F ′ (x) = c ⋅ 1c ec⋅x + 0 = ec⋅x .
So F (x) is an antiderivative of f (x).
Of course, the theorem does not work for c = 0, but then we would have that
f (x) = e0 = 1, which is constant. By the power rule, an antiderivative would be
F (x) = x + C for some constant C.
2. Antiderivative for f (x) =
1
x
We have the power rule for antiderivatives, but it does not work for f (x) = x−1 .
However, we know that the derivative of ln(x) is x1 . So it makes sense that the
antiderivative of x1 should be ln(x). Unfortunately, it is not. But it is close.
Theorem 2.1 (Antiderivative of f (x) = x1 ). Let f (x) = x1 . Then the antiderivatives
of f (x) are of the form F (x) = ln(∣x∣) + C.
Proof. Notice that
F (x) = ln(∣x∣) = {
ln(x) for x > 0
.
ln(−x) for x < 0
For x > 0, we have [ln(x)]′ = x1 .
1
For x < 0, we have from the chain rule that [ln(−x)]′ = − −x
= x1 .
So ln(∣x∣) + C is indeed an antiderivative of f (x) = x1 .
Why do we use F (x) = ln(∣x∣) rather than ln(x)? Notice that ln(x) is defined
only for x > 0, while ln(∣x∣) is defined for all real numbers other than 0. Further
1
2
note that f (x) = x1 is defined also for all real numbers other than 0. It is so the
domains of x1 and its antiderivatives match up.
3. Another definition of ln(x)
We defined ln(x) as the inverse function of ex . There is another way to define
ln(x) using calculus. We define
ln(x) = ∫
x
1
1
dt.
t
In the interest of time, we will not go through the proof that shows our definition
of ln(x) and this new one are exactly the same funtion. But we should be aware
of its existence as it is actually a very historical definition.
4. Integration by substitution
√
Suppose we would like to find ∫
x + 3dx. How could we go about doing this?
√
We need to find a function F (x) whose derivative is x + 3. None of our methods
so far seem to work.
We will make a substitution of variables. The motivation for doing this is to
√ turn
the integral into a more friendly form. Let u = x + 3. Then we would have ∫ udx.
Unfortunately, we cannot take this derivative. Notice that the dx signifies that
we are integrating with respect to x. We need to somehow replace dx by du in
some way. Notice that u is a function of x, so we may differentiate it with respect
to x. du
to the differential
dx = 1. So du = dx. So the differential of u is identical
√
of x, so we can just make the substitution. So we get ∫ udu = 23 u3/2 + C. So
we have taken the derivative, but we would like the function to be in terms of x.
This is simple to fix, just reverse the substitution. Remember, u = x + 3. Thus,
2 3/2
+ C = 23 (x + 3)3/2 + C, which is our antiderivative. We should note that the
3u
C ′ s above are not necessarily the same, but as they are arbitrary numbers, it is
fine to call them both C.
We should check that this is actually√an antiderivative of f (x). We use the chain
rule to get F ′ (x) = 32 ⋅ 32 (x + 3)1/2 ⋅ 1 = x + 3. So, it is indeed an antiderivative.
What we did is a method often called u-subsitution. Intuitively, it “undoes” the
chain rule for derivatives. We know that
d
F (g(x)) = F ′ (g(x)) ⋅ g ′ (x) = f (g(x)) ⋅ g ′ (x),
dx
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where F ′ (x) = f (x). This is just the chain rule. Now integrate both sides to get
d
′
∫ dx F (g(x)) = ∫ f (g(x)) ⋅ g (x).
By the fundamental theorem of calculus, ∫
get
d
dx F (g(x))dx
= F (g(x)) + C, so we
′
∫ f (g(x)) ⋅ g (x)dx = F (g(x)) + C.
What this means is if you have a function of the form f (g(x)) ⋅ g ′ (x) for some
functions f (x) and g(x), then ∫ f (g(x)) ⋅ g ′ (x)dx = F (g(x)) + C, where F (x) is
the antiderivative of f (x). So to use this method, we must watch for functions
that are of the form a composition of functions multiplied by the derivative of the
insider function of that composition. This sounds convoluted, but with practice,
it becomes much more natural.
Example 1. Find ∫ 2x sin(x2 )dx.
Here we have that sin(x2 ) is a composition of functions. If f (x) = sin(x) and
g(x) = x2 , then our function is of the form f (g(x)) ⋅ g ′ (x). So we can use the
substitution method.
We get ∫ sin(x2 ) ⋅ 2xdx = − cos(x2 ) + C. We can easily check this by differentiating.
That was a very nice example, so let’s try something a little harder.
5x6
Example 2. Find ∫
dx.
12x7 + 19
We need a g(x). for simplicity, say g(x) = u = 12x7 + 19. Then
gives du = 84x6 dx. We can easily substitute in u to get
du
dx
= 84x6 , which
5x6
5x6
dx
=
∫ 12x7 + 19
∫ u dx.
We need to somehow get rid of that 5x6 dx and get some sort of du. Well, we
5
know du = 84x6 dx. So 84
du = 5x6 dx. So we have the new integral
5x6
5 1
5
5
7
∫ 12x7 + 19 dx = ∫ 84 u du = 84 ln(∣u∣) + C = 84 ln(∣12x + 19∣) + C.
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We can check this by differentiating with the chain rule.
5. Integration by parts
We have an integration method that “undoes” the chain rule for derivatives.
We now present a method that “undoes” the product rule for derivatives. The
method is know as integration by parts.
From the product rule, we know that
[f (x) ⋅ g(x)]′ = f (x)g ′ (x) + g(x)f ′ (x).
Rearranging, we get
f (x)g ′ (x) = [f (x) ⋅ g(x)]′ − g(x)f ′ (x).
Now we integrate both sides to get
′
′
∫ f (x) ⋅ g (x)dx = f (x) ⋅ g(x) − ∫ g(x) ⋅ f (x)dx.
Finally, call u = f (x) and v = g(x). Then
get the very important formula
du
dx
= f ′ (x) and
dv
dx
= g ′ (x). Then we
∫ u ⋅ dv = u ⋅ v − ∫ v ⋅ du.
Example 3. Find ∫ ln(x)dx.
This may not look like a formula of the form u ⋅ dv, but it is. If we let u = ln(x)
and v = x, then du = x1 dx and dv = 1dx. So
∫ ln(x)dx = ∫ u ⋅ dv = u ⋅ v − ∫ v ⋅ du = ln(x) ⋅ x − ∫ 1dx = ln(x) ⋅ x − x + C.
To reiterate, ∫ ln(x)dx = x ln(x) − x + C.
Example 4. Find ∫ x cos(x)dx.
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The challenge of integration by parts problems is to determine which function
should be u and which should be dv. Let’s see what happens when we let u = cos(x)
and dv = xdx. Then we get v = 21 x2 and du = −sin(x)dx. Then
1 2
1 2
∫ x cos(x)dx = 2 x cos(x) − ∫ − 2 x sin(x)dx.
While this formula is true, our choices of u and dv yielded a much more complicated formula. So let’s switch them. u = x and dv = cos(x)dx. So du = 1dx and
v = sin(x). So we get
∫ x cos(x)dx = x sin(x) − ∫ sin(x)dx = x sin(x) + cos(x) + C.
As with all antiderivative problems, we can check our answer with differentiation.
The integration by parts formula is worth memorizing. Some people use the
mneumonic “ultra-violet voodoo” to help remembering it.