Worked examples
The question
Question: find the integral of 2x times (x² + 1)³.
∫ 2x(x2 + 1)3 dx =
1 Choose u
The inside: x² + 1.
Call it u.
Its derivative is 2x, and 2x is outside.
A good sign: u′ is there, or nearly.
u = x2 + 1
2 Find du
The power rule: du/dx = 2x.
dudx = 2x
Multiply both sides by dx: du = 2x dx.
A small change in u is 2x times a small change in x.
du = 2x dx
3 Write it with u
x² + 1 becomes u, and 2x dx becomes du.
= ∫ u3 du
4 Integrate
Integrate: add 1 to the power, divide by the new power.
3 plus 1 is 4.
= u44 + C
5 Put x back
The question was in x, so the answer is in x too.
Put x² + 1 back in place of u.
So the answer is (x² + 1)⁴ over 4 + C.
= (x2 + 1)44 + C
6 Check
Let's check: take the derivative of our answer with the chain rule.
The power comes down and cancels the number below.
Then times the inside's slope, 2x.
derivative: (x2 + 1)3 × 2x
That is 2x(x² + 1)³: the question again.
Right!
= 2x(x2 + 1)3
Let's look back.
First: choose u.
Then: find du.
Then: write it with u.
Then: integrate.
Then: put x back.
Last: check.
▶ Watch it solved
The question
Question: find the integral of x times (x² + 5)².
∫ x(x2 + 5)2 dx =
1 Choose u
The inside: x² + 5.
Call it u.
Its derivative is 2x, and x is outside.
A good sign: u′ is there, or nearly.
u = x2 + 5
2 Find du
The power rule: du/dx = 2x.
dudx = 2x
Multiply both sides by dx: du = 2x dx.
A small change in u is 2x times a small change in x.
du = 2x dx
We have x dx, not 2x dx.
x is half of 2x.
So x dx is half of du.
x dx = 12 du
3 Write it with u
x² + 5 becomes u, and x dx becomes half of du.
A number in front can wait outside the integral.
= 12∫ u2 du
Careful!
x dx is not du.
It is only half of du.
Careful: not x dx = du ✗
4 Integrate
Integrate: add 1 to the power, divide by the new power.
2 plus 1 is 3.
= 12 × u33 + C
Multiply the numbers: 1/2 times 1/3 is 1/6.
= u36 + C
12 × 13 = 16
5 Put x back
The question was in x, so the answer is in x too.
Put x² + 5 back in place of u.
So the answer is (x² + 5)³ over 6 + C.
= (x2 + 5)36 + C
6 Check
Let's check: take the derivative of our answer with the chain rule.
The power comes down and cancels the number below.
Then times the inside's slope, 2x.
derivative: 12(x2 + 5)2 × 2x
That is x(x² + 5)²: the question again.
Right!
= x(x2 + 5)2
Let's look back.
First: choose u.
Then: find du.
Then: write it with u.
Then: integrate.
Then: put x back.
Last: check.
▶ Watch it solved
The question
Question: find the integral from 0 to 1 of 2x times (x² + 1)³.
∫ from 0 to 1 of 2x(x2 + 1)3 dx =
1 Choose u
The inside: x² + 1.
Call it u.
Its derivative is 2x, and 2x is outside.
A good sign: u′ is there, or nearly.
u = x2 + 1
2 Find du
The power rule: du/dx = 2x.
dudx = 2x
Multiply both sides by dx: du = 2x dx.
A small change in u is 2x times a small change in x.
du = 2x dx
3 Change the limits
The limits 0 and 1 are x values.
u needs its own: x = 0 gives u = 1.
x = 0 → u = 02 + 1 = 0 + 1 = 1
And x = 1 gives u = 2.
x = 1 → u = 12 + 1 = 1 + 1 = 2
4 Write it with u
x² + 1 becomes u, and 2x dx becomes du.
The limits are the u values.
= ∫ from 1 to 2 of u3 du
Careful!
Not the old limits!
0 and 1 are x values, but now we work with u.
Careful: not ∫ from 0 to 1 of u3 du ✗
5 Integrate
Integrate: add 1 to the power, divide by the new power.
3 plus 1 is 4.
= [u44] from 1 to 2
6 Top minus bottom
Put in 2, then take away the same with 1.
= 244 − 144
Work out the powers.
= 164 − 14
So the integral is 15/4.
= 154
7 Check
Let's check without changing the limits: put x back in the antiderivative, and use the x limits 0 and 1.
in x: (x2 + 1)44 from 0 to 1
A calculator gives 4 minus 0.25: 3.75.
The same number.
Right!
4 − 0.25 = 3.75
Let's look back.
First: choose u.
Then: find du.
Then: change the limits.
Then: write it with u.
Then: integrate.
Then: top minus bottom.
Last: check.
▶ Watch it solved
