Worked examples
The question
Question: find the integral of 1, all over x² − 1.
∫ 1x2 − 1 dx =
1 Factor the bottom
First factor the bottom.
x² − 1 is a difference of two squares: x² minus 1 squared.
So it is (x − 1)(x + 1).
x2 − 1 = (x + 1)(x − 1)
a² − b² = (a − b)(a + b), b = 1
2 Split the fraction
Two simple fractions can add up to this one: A over (x + 1), plus B over (x − 1).
We find the numbers A and B.
1(x + 1)(x − 1) = A(x + 1) + B(x − 1)
Careful!
ln of the bottom works only when the top is the bottom's derivative.
Here it is not, so we split.
Careful: not ∫ 1x2 − 1 dx = ln|x2 − 1| ✗
3 Clear the bottoms
Multiply both sides by (x + 1)(x − 1).
On the left the bottom goes.
On the right each fraction keeps the other bracket.
1 = A(x − 1) + B(x + 1)
4 Find A and B
Put in x = −1: it makes (x + 1) zero, so the B part is gone.
x = −1: 1 = A(−2) + B(0)
Divide by −2: A = −1/2.
A = 1−2 = −12
Put in x = 1: it makes (x − 1) zero, so the A part is gone.
x = 1: 1 = B(2) + A(0)
Divide by 2: B = 1/2.
B = 12 = 12
5 Integrate
Now the integral is two easy pieces.
∫ 1(x + 1)(x − 1) dx = ∫ (−12(x + 1) + 12(x − 1)) dx
One over x minus a number integrates to ln of it, like 1 over x.
∫ 1x − p dx = ln|x − p| + C
So the answer is −1/2 × ln|x + 1| + 1/2 × ln|x − 1| + C.
= −12ln|x + 1| + 12ln|x − 1| + C
6 Check
Let's check: add the two fractions back over one bottom, (x + 1)(x − 1).
−12(x + 1) + 12(x − 1) = −12(x − 1) + 12(x + 1)(x + 1)(x − 1)
Open the brackets and add: the top is 1 again.
Right!
= 1(x + 1)(x − 1)
Let's look back.
First: factor the bottom.
Then: split the fraction.
Then: clear the bottoms.
Then: find A and B.
Then: integrate.
Last: check.
Answer = −12ln|x + 1| + 12ln|x − 1| + C
▶ Watch it solved
The question
Question: find the integral of (3x + 1), all over (x − 1)(x + 2).
∫ 3x + 1(x − 1)(x + 2) dx =
1 Split the fraction
Two simple fractions can add up to this one: A over (x − 1), plus B over (x + 2).
We find the numbers A and B.
3x + 1(x − 1)(x + 2) = A(x − 1) + B(x + 2)
Careful!
ln of the bottom works only when the top is the bottom's derivative.
Here it is not, so we split.
Careful: not ∫ 3x + 1(x − 1)(x + 2) dx = ln|(x − 1)(x + 2)| ✗
2 Clear the bottoms
Multiply both sides by (x − 1)(x + 2).
On the left the bottom goes.
On the right each fraction keeps the other bracket.
3x + 1 = A(x + 2) + B(x − 1)
3 Find A and B
Put in x = 1: it makes (x − 1) zero, so the B part is gone.
x = 1: 4 = A(3) + B(0)
Divide by 3: A = 4/3.
A = 43 = 43
Put in x = −2: it makes (x + 2) zero, so the A part is gone.
x = −2: −5 = B(−3) + A(0)
Divide by −3: B = 5/3.
B = −5−3 = 53
4 Integrate
Now the integral is two easy pieces.
∫ 3x + 1(x − 1)(x + 2) dx = ∫ (43(x − 1) + 53(x + 2)) dx
One over x minus a number integrates to ln of it, like 1 over x.
∫ 1x − p dx = ln|x − p| + C
So the answer is 4/3 × ln|x − 1| + 5/3 × ln|x + 2| + C.
= 43ln|x − 1| + 53ln|x + 2| + C
5 Check
Let's check: add the two fractions back over one bottom, (x − 1)(x + 2).
43(x − 1) + 53(x + 2) = 43(x + 2) + 53(x − 1)(x − 1)(x + 2)
Open the brackets and add: the top is 3x + 1 again.
Right!
= 3x + 1(x − 1)(x + 2)
Let's look back.
First: split the fraction.
Then: clear the bottoms.
Then: find A and B.
Then: integrate.
Last: check.
Answer = 43ln|x − 1| + 53ln|x + 2| + C
▶ Watch it solved
The question
Question: find the integral of (4x + 2), all over x² − 2x − 3.
∫ 4x + 2x2 − 2x − 3 dx =
1 Factor the bottom
First factor the bottom.
We need two numbers that multiply to −3 and add to −2: 1 and −3.
x2 − 2x − 3 = (x + 1)(x − 3)
1 × (−3) = −3, 1 + (−3) = −2
2 Split the fraction
Two simple fractions can add up to this one: A over (x + 1), plus B over (x − 3).
We find the numbers A and B.
4x + 2(x + 1)(x − 3) = A(x + 1) + B(x − 3)
Careful!
ln of the bottom works only when the top is the bottom's derivative.
Here it is not, so we split.
Careful: not ∫ 4x + 2x2 − 2x − 3 dx = ln|x2 − 2x − 3| ✗
3 Clear the bottoms
Multiply both sides by (x + 1)(x − 3).
On the left the bottom goes.
On the right each fraction keeps the other bracket.
4x + 2 = A(x − 3) + B(x + 1)
4 Find A and B
Put in x = −1: it makes (x + 1) zero, so the B part is gone.
x = −1: −2 = A(−4) + B(0)
Divide by −4: A = 1/2.
A = −2−4 = 12
Put in x = 3: it makes (x − 3) zero, so the A part is gone.
x = 3: 14 = B(4) + A(0)
Divide by 4: B = 7/2.
B = 144 = 72
5 Integrate
Now the integral is two easy pieces.
∫ 4x + 2(x + 1)(x − 3) dx = ∫ (12(x + 1) + 72(x − 3)) dx
One over x minus a number integrates to ln of it, like 1 over x.
∫ 1x − p dx = ln|x − p| + C
So the answer is 1/2 × ln|x + 1| + 7/2 × ln|x − 3| + C.
= 12ln|x + 1| + 72ln|x − 3| + C
6 Check
Let's check: add the two fractions back over one bottom, (x + 1)(x − 3).
12(x + 1) + 72(x − 3) = 12(x − 3) + 72(x + 1)(x + 1)(x − 3)
Open the brackets and add: the top is 4x + 2 again.
Right!
= 4x + 2(x + 1)(x − 3)
Let's look back.
First: factor the bottom.
Then: split the fraction.
Then: clear the bottoms.
Then: find A and B.
Then: integrate.
Last: check.
Answer = 12ln|x + 1| + 72ln|x − 3| + C
▶ Watch it solved
