Worked examples
The question
Question: find the integral of e³ˣ.
∫ e3x dx =
1 The rules
First, e to the x.
Its derivative is eˣ itself: its slope equals its height.
So its integral is eˣ + C too.
∫ ex dx = ex + C
With e³ˣ: its derivative is 3 times e³ˣ, by the chain rule.
That 3 must not be there.
So we divide by 3.
∫ e3x dx = 13e3x + C
Careful!
The derivative of e³ˣ is 3 times e³ˣ.
So e³ˣ alone is not the answer: divide by 3.
Careful: not ∫ e3x dx = e3x ✗
2 Integrate
Divide by 3: we get 1/3 times e³ˣ.
Add C: the slope cannot tell us a number added on the end.
So the answer is 1/3 × e³ˣ + C.
∫ e3x dx = 13e3x + C
3 Check
Let's check: take the derivative of our answer.
e to the kx gives k times e to the kx.
derivative: 13 × 3e3x
The numbers cancel, and we get e³ˣ: the question again.
Right!
= e3x
Let's look back.
First: the rules.
Then: integrate.
Last: check.
Answer ∫ e3x dx = 13e3x + C
▶ Watch it solved
The question
Question: find the integral of eˣ + 3 over x.
∫ (ex + 3x) dx =
1 The rules
First, e to the x.
Its derivative is eˣ itself: its slope equals its height.
So its integral is eˣ + C too.
∫ ex dx = ex + C
Now 1 over x.
The power rule cannot do it: x to the −1 would become x to the 0, over 0.
Instead: the derivative of ln x is 1 over x.
So the integral of 1/x is ln|x| + C.
∫ 1x dx = ln|x| + C
The bars mean the size of x.
ln only takes positive numbers, and x may be negative.
|x|: ln needs a number > 0
Careful!
The power rule fails for 1/x: the new power would be 0, and we cannot divide by 0.
Use ln.
Careful: not x00 ✗
2 Each term
e to the x stays e to the x.
∫ ex dx = ex
3 times 1 over the bottom gives 3 ln|x|.
∫ 3x dx = 3ln|x|
Put the terms together, and add C: the slope cannot tell us a number added on the end.
So the answer is eˣ + 3 ln|x| + C.
= ex + 3ln|x| + C
3 Check
Let's check: take the derivative of our answer.
ln of the bottom gives its slope over the bottom.
derivative: ex + 3 × 1x
The numbers cancel, and we get eˣ + 3/x: the question again.
Right!
= ex + 3x
Let's look back.
First: the rules.
Then: each term.
Last: check.
▶ Watch it solved
The question
Question: find the integral of 8e⁻²ˣ + 3/x − 6x².
∫ (8e−2x + 3x − 6x2) dx =
1 The rules
First, e to the x.
Its derivative is eˣ itself: its slope equals its height.
So its integral is eˣ + C too.
∫ ex dx = ex + C
With e⁻²ˣ: its derivative is −2 times e⁻²ˣ, by the chain rule.
That −2 must not be there.
So we divide by −2.
∫ e−2x dx = 1−2e−2x + C
Now 1 over x.
The power rule cannot do it: x to the −1 would become x to the 0, over 0.
Instead: the derivative of ln x is 1 over x.
So the integral of 1/x is ln|x| + C.
∫ 1x dx = ln|x| + C
The bars mean the size of x.
ln only takes positive numbers, and x may be negative.
|x|: ln needs a number > 0
Powers of x use the power rule: add 1 to the power, divide by the new power.
xn → xn + 1n + 1
Careful!
The derivative of e⁻²ˣ is −2 times e⁻²ˣ.
So e⁻²ˣ alone is not the answer: divide by −2.
Careful: not ∫ e−2x dx = e−2x ✗
2 Each term
Divide by −2: 8 over −2 is −4.
∫ 8e−2x dx = −4e−2x
3 times 1 over the bottom gives 3 ln|x|.
∫ 3x dx = 3ln|x|
The power rule.
∫ −6x2 dx = −2x3
Put the terms together, and add C: the slope cannot tell us a number added on the end.
So the answer is −4e⁻²ˣ + 3 ln|x| − 2x³ + C.
= −4e−2x + 3ln|x| − 2x3 + C
3 Check
Let's check: take the derivative of our answer.
e to the kx gives k times e to the kx.
ln of the bottom gives its slope over the bottom.
derivative: −4 × −2e−2x + 3 × 1x − 6x2
The numbers cancel, and we get 8e⁻²ˣ + 3/x − 6x²: the question again.
Right!
= 8e−2x + 3x − 6x2
Let's look back.
First: the rules.
Then: each term.
Last: check.
Answer = −4e−2x + 3ln|x| − 2x3 + C
▶ Watch it solved
