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).

1 c⋅x ′ 1 c⋅x c⋅x Proof. Consider F (x) = c e +D. Then by the chain rule, F (x) = c⋅ c e +0 = e . 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.

1 2. Antiderivative for f(x) = x We have the power rule for antiderivatives, but it does not work for f(x) = x−1. 1 However, we know that the derivative of ln(x) is x . So it makes sense that the 1 antiderivative of x should be ln(x). Unfortunately, it is not. But it is close.

1 1 Theorem 2.1 (Antiderivative of f(x) = x ). Let f(x) = x . Then the antiderivatives of f(x) are of the form F (x) = ln(SxS) + C.

Proof. Notice that

ln(x) for x > 0 F (x) = ln(SxS) = œ . ln(−x) for x < 0

′ 1 For x > 0, we have [ln(x)] = x .

′ 1 1 For x < 0, we have from the chain rule that [ln(−x)] = − −x = x .

1 So ln(SxS) + C is indeed an antiderivative of f(x) = x . 

Why do we use F (x) = ln(SxS) rather than ln(x)? Notice that ln(x) is defined only for x > 0, while ln(SxS) is defined for all real numbers other than 0. Further 1 2

1 note that f(x) = x is defined also for all real numbers other than 0. It is so the 1 domains of x 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 x 1 ln(x) = dt. S1 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 S 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 du to x. dx = 1. So du = dx. So the differential of u is identical to the differential of x, so we can just make the substitution. So we get √udu 2 u3~2 C. So ∫ = 3 + 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 2 3~2 3 u + C = 3 (x + 3) + C, which is our antiderivative. We should note that the 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 ′ 3 2 1~2 √ rule to get F (x) = 2 ⋅ 3 (x + 3) ⋅ 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 ( ( )) = ( ( )) ⋅ ( ) = ( ( )) ⋅ ( ) 3

where F ′(x) = f(x). This is just the chain rule. Now integrate both sides to get d F g x f g x g′ x . S dx ( ( )) = S ( ( )) ⋅ ( ) By the fundamental theorem of calculus, d F g x dx F g x C, so we ∫ dx ( ( )) = ( ( )) + get

′ S 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 S 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.

2 Example 1. Find ∫ 2x sin(x )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.

2 2 We get ∫ sin(x ) ⋅ 2xdx = − cos(x ) + C. We can easily check this by differenti- ating.

That was a very nice example, so let’s try something a little harder. 5x6 Example 2. Find dx. S 12x7 + 19

7 du 6 We need a g(x). for simplicity, say g(x) = u = 12x + 19. Then dx = 84x , which gives du = 84x6dx. We can easily substitute in u to get 5x6 5x6 dx = dx. S 12x7 + 19 S u We need to somehow get rid of that 5x6dx and get some sort of du. Well, we 6 5 6 know du = 84x dx. So 84 du = 5x dx. So we have the new integral

6 5x 5 1 5 5 7 dx = du = ln(SuS) + C = ln(S12x + 19S) + C. S 12x7 + 19 S 84 u 84 84 4

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

′ ′ S f(x) ⋅ g (x)dx = f(x) ⋅ g(x) − S g(x) ⋅ f (x)dx.

du ′ dv ′ Finally, call u = f(x) and v = g(x). Then dx = f (x) and dx = g (x). Then we get the very important formula

S u ⋅ dv = u ⋅ v − S 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) 1 and v = x, then du = x dx and dv = 1dx. So

S ln(x)dx = S u ⋅ dv = u ⋅ v − S v ⋅ du = ln(x) ⋅ x − S 1dx = ln(x) ⋅ x − x + C.

To reiterate, ∫ ln(x)dx = x ln(x) − x + C.

Example 4. Find ∫ x cos(x)dx. 5

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) 1 2 and dv = xdx. Then we get v = 2 x and du = −sin(x)dx. Then 1 1 x cos x dx x2 cos x x2 sin x dx. S ( ) = 2 ( ) − S −2 ( ) While this formula is true, our choices of u and dv yielded a much more com- plicated formula. So let’s switch them. u = x and dv = cos(x)dx. So du = 1dx and v = sin(x). So we get

S x cos(x)dx = x sin(x) − S 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.